jshbtchr jshbtchr - 2 months ago 9
Python Question

Conditional in Django Tables

I am trying to include a conditional in my Django table, but I am having some issues finding the correct syntax to do so.

I have a boolean field in one of my models, and based on that value - I would like to render a different

label_link
and callback url for that specific db record.

Here is the code I am using:

class Feature(BaseTable):
orderable = False
title_english = tables.Column()
featured_item = tables.Column()
if (featured_item == False):
actions = tables.TemplateColumn("""
{% url 'add_feature' pk=record.pk as url_ %}
{% label_link url_ 'Add to Features' %}
""", attrs=dict(cell={'class': 'span1'}))
else:
actions = tables.TemplateColumn("""
{% url 'remove_feature' pk=record.pk as url_ %}
{% label_link url_ 'Remove From Features' %}
""", attrs=dict(cell={'class': 'span1'}))


Now I am aware currently that this is just checking to see if the value exists - therefore always rendering the code under the
else
statement.

I have been unable to find any documentation that covers this particular nuance of Django tables.

Note: I'm using Django 1.5 and Python 2.7

Answer

You can't put the if statement in the Feature class - it is processed when the class is loaded, so you don't have access to the table data yet.

Inside the TemplateColumn, you can access the current row of the table with record, so you could move the logic there.

class Feature(BaseTable):
    ...
    actions = tables.TemplateColumn("""
    {% if not record.featured_item %}
    {% url 'add_feature' pk=record.pk as url_ %}
    {% label_link url_ 'Add to Features' %}
    {% else %}
    {% url 'remove_feature' pk=record.pk as url_ %}
    {% label_link url_ 'Remove From Features' %}
    {% endif %}
    """, attrs=dict(cell={'class': 'span1'}))
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