How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

unboundlocalerror local variable referenced before assignment django

One error you might encounter when running Python code is:

This error commonly occurs when you reference a variable inside a function without first assigning it a value.

You could also see this error when you forget to pass the variable as an argument to your function.

Let me show you an example that causes this error and how I fix it in practice.

How to reproduce this error

Suppose you have a variable called name declared in your Python code as follows:

Next, you created a function that uses the name variable as shown below:

When you execute the code above, you’ll get this error:

This error occurs because you both assign and reference a variable called name inside the function.

Python thinks you’re trying to assign the local variable name to name , which is not the case here because the original name variable we declared is a global variable.

How to fix this error

To resolve this error, you can change the variable’s name inside the function to something else. For example, name_with_title should work:

As an alternative, you can specify a name parameter in the greet() function to indicate that you require a variable to be passed to the function.

When calling the function, you need to pass a variable as follows:

This code allows Python to know that you intend to use the name variable which is passed as an argument to the function as part of the newly declared name variable.

Still, I would say that you need to use a different name when declaring a variable inside the function. Using the same name might confuse you in the future.

Here’s the best solution to the error:

Now it’s clear that we’re using the name variable given to the function as part of the value assigned to name_with_title . Way to go!

The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable.

To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function.

I hope this tutorial is useful. See you in other tutorials.

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local variable 'nextofkin' referenced before assignment

Hi! Everything was working fine until when i added the try and catch block, this ‘unboundLocalErro’ occurs. I check whether the variable tally with my models objects but not. I checked my indentation but everything is ok. Please I need help.

If this exception is triggered, there is no assignment to a variable named nextofkin . Therefore, there is no variable by that name when you enter the if request.method ... block.

Ok. So what is the solution? Because i need that exception.

The solution depends, on what you want to achieve. Normally in this scenario your render the template again, redirect the user. But if you still want to keep processing with this, you may set nextofkin to None on the except block.

Thanks. It worked! I set the nextofkin variable to None

Related Topics

Fixing Python UnboundLocalError: Local Variable ‘x’ Accessed Before Assignment

Understanding unboundlocalerror.

The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope. This error typically surfaces when using a variable that has not been initialized in the current function’s scope or when an attempt is made to modify a global variable without proper declaration.

Solutions for the Problem

To fix an UnboundLocalError, you need to identify the scope of the problematic variable and ensure it is correctly used within that scope.

Method 1: Initializing the Variable

Make sure to initialize the variable within the function before using it. This is often the simplest fix.

Method 2: Using Global Variables

If you intend to use a global variable and modify its value within a function, you must declare it as global before you use it.

Method 3: Using Nonlocal Variables

If the variable is defined in an outer function and you want to modify it within a nested function, use the nonlocal keyword.

That’s it. Happy coding!

Next Article: Fixing Python TypeError: Descriptor 'lower' for 'str' Objects Doesn't Apply to 'dict' Object

Previous Article: Fixing Python NameError – name ‘List’ is not defined

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Local variable referenced before assignment in Python

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Last updated: Feb 17, 2023 Reading time · 4 min

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# Local variable referenced before assignment in Python

The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.

To solve the error, mark the variable as global in the function definition, e.g. global my_var .

unboundlocalerror local variable name referenced before assignment

Here is an example of how the error occurs.

We assign a value to the name variable in the function.

# Mark the variable as global to solve the error

To solve the error, mark the variable as global in your function definition.

mark variable as global

If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .

# Local variables shadow global ones with the same name

You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.

accessing global variables in functions

Accessing the name variable in the function is perfectly fine.

On the other hand, variables declared in a function cannot be accessed from the global scope.

variables declared in function cannot be accessed in global scope

The name variable is declared in the function, so trying to access it from outside causes an error.

Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.

# Returning a value from the function instead

An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.

return value from the function

We simply return the value that we eventually use to assign to the name global variable.

# Passing the global variable as an argument to the function

You should also consider passing the global variable as an argument to the function.

pass global variable as argument to function

We passed the name global variable as an argument to the function.

If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global .

# Assigning a value to a local variable from an outer scope

If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.

assign value to local variable from outer scope

The nonlocal keyword allows us to work with the local variables of enclosing functions.

Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.

not using nonlocal prints empty string

Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.

Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.

Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.

# Discussion

As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:

  • Becomes local to the scope.
  • Shadows any variables from the outer scope that have the same name.

The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.

At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.

The most intuitive way to solve the error is to use the global keyword.

The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.

  • If a variable is only referenced inside a function, it is implicitly global.
  • If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .

If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

  • SyntaxError: name 'X' is used prior to global declaration

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[SOLVED] Local Variable Referenced Before Assignment

local variable referenced before assignment

Python treats variables referenced only inside a function as global variables. Any variable assigned to a function’s body is assumed to be a local variable unless explicitly declared as global.

Why Does This Error Occur?

Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.

Before we hop into the solutions, let’s have a look at what is the global and local variables.

Local Variable Declarations vs. Global Variable Declarations

[Fixed] typeerror can’t compare datetime.datetime to datetime.date

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Local Variable Referenced Before Assignment Error

Explanation

The variable myVar has been assigned a value twice. Once before the declaration of myFunction and within myFunction itself.

Using Global Variables

Passing the variable as global allows the function to recognize the variable outside the function.

Create Functions that Take in Parameters

Instead of initializing myVar as a global or local variable, it can be passed to the function as a parameter. This removes the need to create a variable in memory.

UnboundLocalError: local variable ‘DISTRO_NAME’

This error may occur when trying to launch the Anaconda Navigator in Linux Systems.

Upon launching Anaconda Navigator, the opening screen freezes and doesn’t proceed to load.

Try and update your Anaconda Navigator with the following command.

If solution one doesn’t work, you have to edit a file located at

After finding and opening the Python file, make the following changes:

In the function on line 159, simply add the line:

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

The program takes information from a form filled out by a user. Accordingly, an email is sent using the information.

Upon running you get the following error:

We have created a class myForm that creates instances of Django forms. It extracts the user’s name, email, and message to be sent.

A function GetContact is created to use the information from the Django form and produce an email. It takes one request parameter. Prior to sending the email, the function verifies the validity of the form. Upon True , .get() function is passed to fetch the name, email, and message. Finally, the email sent via the send_mail function

Why does the error occur?

We are initializing form under the if request.method == “POST” condition statement. Using the GET request, our variable form doesn’t get defined.

Local variable Referenced before assignment but it is global

This is a common error that happens when we don’t provide a value to a variable and reference it. This can happen with local variables. Global variables can’t be assigned.

This error message is raised when a variable is referenced before it has been assigned a value within the local scope of a function, even though it is a global variable.

Here’s an example to help illustrate the problem:

In this example, x is a global variable that is defined outside of the function my_func(). However, when we try to print the value of x inside the function, we get a UnboundLocalError with the message “local variable ‘x’ referenced before assignment”.

This is because the += operator implicitly creates a local variable within the function’s scope, which shadows the global variable of the same name. Since we’re trying to access the value of x before it’s been assigned a value within the local scope, the interpreter raises an error.

To fix this, you can use the global keyword to explicitly refer to the global variable within the function’s scope:

However, in the above example, the global keyword tells Python that we want to modify the value of the global variable x, rather than creating a new local variable. This allows us to access and modify the global variable within the function’s scope, without causing any errors.

Local variable ‘version’ referenced before assignment ubuntu-drivers

This error occurs with Ubuntu version drivers. To solve this error, you can re-specify the version information and give a split as 2 –

Here, p_name means package name.

With the help of the threading module, you can avoid using global variables in multi-threading. Make sure you lock and release your threads correctly to avoid the race condition.

When a variable that is created locally is called before assigning, it results in Unbound Local Error in Python. The interpreter can’t track the variable.

Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.

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[Fixed] nameerror: name Unicode is not defined

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Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

If you try to reference a local variable before assigning a value to it within the body of a function, you will encounter the UnboundLocalError: local variable referenced before assignment.

The preferable way to solve this error is to pass parameters to your function, for example:

Alternatively, you can declare the variable as global to access it while inside a function. For example,

This tutorial will go through the error in detail and how to solve it with code examples .

Table of contents

What is scope in python, unboundlocalerror: local variable referenced before assignment, solution #1: passing parameters to the function, solution #2: use global keyword, solution #1: include else statement, solution #2: use global keyword.

Scope refers to a variable being only available inside the region where it was created. A variable created inside a function belongs to the local scope of that function, and we can only use that variable inside that function.

A variable created in the main body of the Python code is a global variable and belongs to the global scope. Global variables are available within any scope, global and local.

UnboundLocalError occurs when we try to modify a variable defined as local before creating it. If we only need to read a variable within a function, we can do so without using the global keyword. Consider the following example that demonstrates a variable var created with global scope and accessed from test_func :

If we try to assign a value to var within test_func , the Python interpreter will raise the UnboundLocalError:

This error occurs because when we make an assignment to a variable in a scope, that variable becomes local to that scope and overrides any variable with the same name in the global or outer scope.

var +=1 is similar to var = var + 1 , therefore the Python interpreter should first read var , perform the addition and assign the value back to var .

var is a variable local to test_func , so the variable is read or referenced before we have assigned it. As a result, the Python interpreter raises the UnboundLocalError.

Example #1: Accessing a Local Variable

Let’s look at an example where we define a global variable number. We will use the increment_func to increase the numerical value of number by 1.

Let’s run the code to see what happens:

The error occurs because we tried to read a local variable before assigning a value to it.

We can solve this error by passing a parameter to increment_func . This solution is the preferred approach. Typically Python developers avoid declaring global variables unless they are necessary. Let’s look at the revised code:

We have assigned a value to number and passed it to the increment_func , which will resolve the UnboundLocalError. Let’s run the code to see the result:

We successfully printed the value to the console.

We also can solve this error by using the global keyword. The global statement tells the Python interpreter that inside increment_func , the variable number is a global variable even if we assign to it in increment_func . Let’s look at the revised code:

Let’s run the code to see the result:

Example #2: Function with if-elif statements

Let’s look at an example where we collect a score from a player of a game to rank their level of expertise. The variable we will use is called score and the calculate_level function takes in score as a parameter and returns a string containing the player’s level .

In the above code, we have a series of if-elif statements for assigning a string to the level variable. Let’s run the code to see what happens:

The error occurs because we input a score equal to 40 . The conditional statements in the function do not account for a value below 55 , therefore when we call the calculate_level function, Python will attempt to return level without any value assigned to it.

We can solve this error by completing the set of conditions with an else statement. The else statement will provide an assignment to level for all scores lower than 55 . Let’s look at the revised code:

In the above code, all scores below 55 are given the beginner level. Let’s run the code to see what happens:

We can also create a global variable level and then use the global keyword inside calculate_level . Using the global keyword will ensure that the variable is available in the local scope of the calculate_level function. Let’s look at the revised code.

In the above code, we put the global statement inside the function and at the beginning. Note that the “default” value of level is beginner and we do not include the else statement in the function. Let’s run the code to see the result:

Congratulations on reading to the end of this tutorial! The UnboundLocalError: local variable referenced before assignment occurs when you try to reference a local variable before assigning a value to it. Preferably, you can solve this error by passing parameters to your function. Alternatively, you can use the global keyword.

If you have if-elif statements in your code where you assign a value to a local variable and do not account for all outcomes, you may encounter this error. In which case, you must include an else statement to account for the missing outcome.

For further reading on Python code blocks and structure, go to the article: How to Solve Python IndentationError: unindent does not match any outer indentation level .

Go to the  online courses page on Python  to learn more about Python for data science and machine learning.

Have fun and happy researching!

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Opened 11 years ago

Closed 11 years ago

Last modified 11 years ago

#20761 closed Bug ( fixed )

Unboundlocalerror: local variable 'sid' referenced before assignment when using get_or_create with a badly formed __init__, description.

The cause of this bug seems to be a little convoluted. I don't expect it to happen in normal use, but it still seems to be unexpected behaviour.

The smallest testcase I can create it with is:

The cause of the Integrity error isn't important. I was trying some stuff in a custom init that didn't end up working. The error I get is:

The reason for this seems to be that the try: block in query.py:get_or_create expects the IntegrityError to be thrown when the new object is saved, not when it's created (line 476).

Change History (8)

Comment:1 by claude paroz , 11 years ago.

I guess that the obj = self.model(**params) line (now in _create_object_from_params in master) should be outside the try/except construct.

comment:2 by loic84 , 11 years ago

@claudep the try/except construct is there to catch a race condition.

[0e51d8eb66121b2558a38c9f4e366df781046873] changed the method to handle transaction better, however I don't like how any error would lead to a self.get() .

I'd suggest the following snippet:

Refs #20463 #20429 #12579 #13906 .

Edit: Probably shouldn't look at the ORM code at 4AM. The self.model(**params) part can/should be moved outside of the try/except construct, only the save logic needs to be there.

comment:3 by loic84 , 11 years ago

PR ​ https://github.com/django/django/pull/1385

comment:4 by loic84 , 11 years ago

Comment:5 by tim graham <timograham@…> , 11 years ago.

In e716518ad29898fb25c820023aaf2fdd1c9eb4af :

Fixed #20761 -- Fixed DatabaseError handling in get_or_create and update_or_create.

comment:6 by Tim Graham <timograham@…> , 11 years ago

In ad98b985aa18fbca65b45aae2421b985904bf561 :

Fixed test failures introduced in e716518ad29898fb25c820023aaf2fdd1c9eb4af

refs #20761

comment:7 by Claude Paroz <claude@…> , 11 years ago

In 311c1d2848bde59bf03627e5c3d72b319285201b :

Fixed #20761 -- Reworded ForeignKey default error message

comment:8 by Claude Paroz , 11 years ago

Oups, sorry for the wrong ticket number, the latest commit above refers to #20791

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Python local variable referenced before assignment Solution

When you start introducing functions into your code, you’re bound to encounter an UnboundLocalError at some point. This error is raised when you try to use a variable before it has been assigned in the local context .

In this guide, we talk about what this error means and why it is raised. We walk through an example of this error in action to help you understand how you can solve it.

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What is unboundlocalerror: local variable referenced before assignment.

Trying to assign a value to a variable that does not have local scope can result in this error:

Python has a simple rule to determine the scope of a variable. If a variable is assigned in a function , that variable is local. This is because it is assumed that when you define a variable inside a function you only need to access it inside that function.

There are two variable scopes in Python: local and global. Global variables are accessible throughout an entire program; local variables are only accessible within the function in which they are originally defined.

Let’s take a look at how to solve this error.

An Example Scenario

We’re going to write a program that calculates the grade a student has earned in class.

We start by declaring two variables:

These variables store the numerical and letter grades a student has earned, respectively. By default, the value of “letter” is “F”. Next, we write a function that calculates a student’s letter grade based on their numerical grade using an “if” statement :

Finally, we call our function:

This line of code prints out the value returned by the calculate_grade() function to the console. We pass through one parameter into our function: numerical. This is the numerical value of the grade a student has earned.

Let’s run our code and see what happens:

An error has been raised.

The Solution

Our code returns an error because we reference “letter” before we assign it.

We have set the value of “numerical” to 42. Our if statement does not set a value for any grade over 50. This means that when we call our calculate_grade() function, our return statement does not know the value to which we are referring.

We do define “letter” at the start of our program. However, we define it in the global context. Python treats “return letter” as trying to return a local variable called “letter”, not a global variable.

We solve this problem in two ways. First, we can add an else statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

If you are using an “if” statement where you declare a variable, you should make sure there is an “else” statement in place. This will make sure that even if none of your if statements evaluate to True, you can still set a value for the variable with which you are going to work.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our calculate_grade() function. However, this approach is likely to lead to more confusing code and other issues. In general, variables should not be declared using “global” unless absolutely necessary . Your first, and main, port of call should always be to make sure that a variable is correctly defined.

In the example above, for instance, we did not check that the variable “letter” was defined in all use cases.

That’s it! We have fixed the local variable error in our code.

The UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. You can solve this error by ensuring that a local variable is declared before you assign it a value.

Now you’re ready to solve UnboundLocalError Python errors like a professional developer !

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UnboundLocalError: local variable 'field' referenced before assignment #68

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COMMENTS

  1. UnboundLocalError- local variable referenced before assignment

    1 Answer Sorted by: 8 This is happening because request.method == 'POST' is True (first condition passes) but form.is_valid () is False (second nested condition fails), which means the final return after the else is run but ctxt is not defined. Perhaps you intended that final return to be indented as a part of the else clause? Share

  2. local variable 'form' referenced before assignment

    your are getting local variable 'form' referenced before assignment this error for this line of code: context = { 'form': form, 'all_data': total_data }

  3. How to fix UnboundLocalError: local variable 'x' referenced before

    The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable. To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function. I hope this tutorial is useful.

  4. local variable 'nextofkin' referenced before assignment

    Hi! Everything was working fine until when i added the try and catch block, this 'unboundLocalErro' occurs. I check whether the variable tally with my models objects but not. I checked my indentation but everything is ok. Please I need help. def nextofkinupdate(request, id): try: nextofkin = Next_of_kin.objects.get(user_id = id)

  5. Fixing Python UnboundLocalError: Local Variable 'x' Accessed Before

    The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope.

  6. Local variable referenced before assignment in Python

    If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global. # Local variables shadow global ones with the same name You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.

  7. [SOLVED] Local Variable Referenced Before Assignment

    Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.

  8. Python UnboundLocalError: local variable referenced before assignment

    UnboundLocalError: local variable referenced before assignment. Example #1: Accessing a Local Variable. Solution #1: Passing Parameters to the Function. Solution #2: Use Global Keyword. Example #2: Function with if-elif statements. Solution #1: Include else statement. Solution #2: Use global keyword. Summary.

  9. #7151 (Management UnBoundLocalError error on local variable ...

    UnboundLocalError: local variable 'args' referenced before assignment Because the preprocessor tries to extract --settings and --pythonpath from an invalid command line, thus raising an exception which is caught by the code and ignored. (See full error message below)

  10. UnboundLocalError: local variable 'sid' referenced before ...

    UnboundLocalError: local variable 'sid' referenced before assignment when using get_or_create with a badly formed __init__ Description ¶ The cause of this bug seems to be a little convoluted. I don't expect it to happen in normal use, but it still seems to be unexpected behaviour. The smallest testcase I can create it with is:

  11. Python local variable referenced before assignment Solution

    Trying to assign a value to a variable that does not have local scope can result in this error: UnboundLocalError: local variable referenced before assignment. Python has a simple rule to determine the scope of a variable. If a variable is assigned in a function, that variable is local. This is because it is assumed that when you define a ...

  12. UnboundLocalError: local variable 'key_cls' referenced before ...

    @zout this traceback is useless without a code snippet to reproduce the problem. As you can see from our test suite we've added some code samples for the use-cases we could think of. When you see a failure for something that apparently works in CI you have to provide us with a reproducer.

  13. UnboundLocalError: local variable 'field' referenced before assignment

    Saved searches Use saved searches to filter your results more quickly

  14. Issue loading Django fixture. UnboundLocalError: local variable 'pk

    UnboundLocalError: local variable 'pk' referenced before assignment I have the following models for which I will attempt to load data using a Django fixture file. class StaffPosition(models.Model):

  15. r/django on Reddit: UnboundLocalError with an unassigned local variable

    Hey fellow djangonians. I have been following the django docs (and et al) while trying to build a web app. What I want to accomplish is, create a form from an existing model.

  16. Python 3: UnboundLocalError: local variable referenced before assignment

    The following code gives the error UnboundLocalError: local variable 'Var1' referenced before assignment: Var1 = 1 Var2 = 0 def function (): if Var2 == 0 and Var1 > 0: print ("Result 1") elif Var2 == 1 and Var1 > 0: print ("Result 2") elif Var1 < 1: print ("Result 3") Var1 -= 1 function () How can I fix this? python python-3.x scope Share

  17. UnboundLocalError: local variable 'form' referenced before assignment

    There are a lot of unnecessary lines in your code, but anyway... Your problem is that you are importing the forms.py module as forms, but then you are referencing it as form (without the 's' at the end) when trying to instantiate the form class. Your view should look like:

  18. UnboundLocalError local variable 'context' referenced before assignment

    UnboundLocalError at /blogapp/ local variable 'context' referenced before assignment Request Method: GET Request URL: http://127...1:8000/blogapp/ Django Version: 4.0.2 Exception Type: UnboundLocalError Exception Value: local variable 'context' referenced before assignment My views

  19. UnboundLocalError: local variable referenced before assignment

    It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment whic... Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.