CODEWITHSUNDEEP
pythondjango
changzhi
changzhi

Reputation: 2709

Django template "with" tag invoke an variable

There is a Django template, I want to use instance.links[0]['href'].split(':')[1] as a variable, and use a with tag to invoke it:

{% if instance %}
{% with "http:{{ instance.links[0]['href'].split(':')[1] }}/dashboard/speed?speed=" as url%}
{% endwith %}
{% endif %}

And this is instance:

def set_network_speed(instance):

    template_name = 'project/instances/set_network_speed.html'
    context = {"instance": instance}
    return template.loader.render_to_string(template_name, context)

The above code is wrong. Could someone help me to fix it? Thanks a lot !

Upvotes: 0

Views: 97

Answers (2)

christophe31
christophe31

Reputation: 6467

The best way for you is to add an empty args method on your instance object class.

class InstanceClass(AnyThing):
    def get_url(self):
        return "http://" + instance.links[0]['href'].split(':')[1] + "/dashboard/speed?speed="

Then you can use it in the template

{% if instance %}
{% with instance_url=instance.get_url %}
{% endwith %}
{% endif %}

edit: Thanx for the comment Quentin Pradet. Fix can be implemented in the question too.

Even if it will certainly not conflict in templates, I would not use an existing command name as var name to avoid confusion.

Upvotes: 1

Daniel Roseman
Daniel Roseman

Reputation: 599778

No. Your can't do that in a template. Even if you did the rest of the syntax, you can't pass an argument to split.

You will need to write a custom template tag or filter.

Upvotes: 1

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