Template permission checks in Django - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What will this Django template render?

Given the following Django template snippet, what will be rendered if the user has the 'app.view_item' permission?

{% if perms.app.view_item %}You can view this item.{% else %}Access denied.{% endif %}

Django

{% if perms.app.view_item %}You can view this item.{% else %}Access denied.{% endif %}

A You can view this item.

B Access denied.

C An error occurs because 'perms' is undefined.

D Nothing is rendered.

💡 Hint

The 'perms' variable in Django templates checks user permissions.

The 'perms' context variable allows checking if the user has a specific permission. If the user has 'app.view_item', the first block renders.

📝 Syntax

intermediate

Identify the syntax error in this permission check

What is wrong with this Django template code?

{% if perms.app.view_item == True %}Allowed{% else %}Denied{% endif %}

Django

{% if perms.app.view_item == True %}Allowed{% else %}Denied{% endif %}

A The comparison '== True' is unnecessary and causes a syntax error.

B The 'perms' variable cannot be used in templates.

C The 'else' tag is missing a colon.

D The template tag should be {% if perms.app.view_item %} without '== True'.

💡 Hint

Django template if tags do not require explicit boolean comparisons.

In Django templates, 'perms.app.view_item' already evaluates to True or False. Using '== True' is not a syntax error but unnecessary. The correct style is without '== True'.

🔧 Debug

advanced

Why does this permission check always fail?

Consider this Django template snippet:

{% if perms.app.view_item %}Show{% else %}Hide{% endif %}

The user has the permission 'app.view_item', but the template always renders 'Hide'. Why?

Django

{% if perms.app.view_item %}Show{% else %}Hide{% endif %}

A The user object is not passed to the template context, so 'perms' is empty.

B The permission name is incorrect; it should be 'app.viewitems'.

C The template tag syntax is invalid and causes the condition to always be false.

D The user is anonymous, so no permissions are granted.

💡 Hint

Check if the user context is available in the template.

If the user is not passed to the template context or the authentication context processor is missing, 'perms' will not have permission info, causing the check to fail.

❓ state_output

advanced

What is the output of this permission loop?

Given this Django template code:

{% for perm in perms.app %}{{ perm }} {% endfor %}

What will be the output if the user has permissions 'app.view_item' and 'app.change_item'?

Django

{% for perm in perms.app %}{{ perm }} {% endfor %}

A Outputs 'view_item change_item '

B Outputs nothing because 'perms.app' is empty.

C Raises an error because 'perms.app' is not iterable.

D Outputs 'app.view_item app.change_item '

💡 Hint

'perms.app' is not a list but a permission checker object.

'perms.app' is not iterable in Django templates. It is a special object to check permissions, so looping over it causes an error.

🧠 Conceptual

expert

Which template code correctly checks multiple permissions?

You want to show a message only if the user has both 'app.view_item' and 'app.change_item' permissions. Which Django template snippet does this correctly?

A {% if perms.app.view_item %}{% if perms.app.change_item %}Allowed{% else %}Denied{% endif %}{% else %}Denied{% endif %}

B {% if perms.app.view_item and perms.app.change_item %}Allowed{% else %}Denied{% endif %}

C {% if perms.app.view_item or perms.app.change_item %}Allowed{% else %}Denied{% endif %}

D {% if perms.app.view_item & perms.app.change_item %}Allowed{% else %}Denied{% endif %}

💡 Hint

Use 'and' to combine conditions in Django templates.

Django templates support 'and' for combining conditions. Option B uses 'and' correctly. Option B is nested but more complex. Option B uses 'or' which is incorrect. Option B uses '&' which is invalid in templates.