Use variable in url_for in flask for parameter name

Question

I have this code in my macro file for flask.

{% macro display_name(name, parameter, parameter_value) %}
<a href="{{ url_for('users_stats',parameter = parameter_value) }}">{{name}}</a>

{% endmacro %}

I am not able to set the value of parameter to what I pass in the function. It is coming as a plain text, How to fix this?

This is how I am calling the function.

  {{ display_name("Rahul", 'age', 15) }}

So you want to assign parameter `age` to value `15` and call `url_for`? — mikulatomas, Jan 28 '22 at 10:17
Yes, I want to assign age to 'parameter' and 15 to parameter_value — Ansh, Jan 28 '22 at 10:19
Possibly related https://stackoverflow.com/questions/47907268/how-do-i-pass-a-string-as-an-argument-name you can use kwargs and dict unpacking — mikulatomas, Jan 28 '22 at 10:20

score 0 · Accepted Answer · answered Jan 28 '22 at 10:26

Try kwargs dict unpacking. I recommend research more info later.

{% macro display_name(name, parameter, parameter_value) %}
<a href="{{ url_for('users_stats', **{parameter: parameter_value}) }}">{{name}}</a>

{% endmacro %}

The point is creating dict where keys are names of parameters, values are values of corresponding parameters. Then unpack this dict via ** operator to pass these parameters and values into function.

Use variable in url_for in flask for parameter name

1 Answers1