How to output a comma delimited list in jinja python template?

Michael Zippo

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If I have a list of users say ["Sam", "Bob", "Joe"], I want to do something where I can output in my jinja template file:

{% for user in userlist %}
    <a href="/profile/{{ user }}/">{{ user }}</a>
    {% if !loop.last %}
        , 
    {% endif %}
{% endfor %}

I want to make the output template be:

Sam, Bob, Joe

I tried the above code to check if it was on the last iteration of the loop and if not, then don"t insert a comma, but it does not work. How do I do this?

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I would like to know how to put a time delay in a Python script.

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Answer #1

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
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Answer #2

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

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os.remove() removes a file.
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Path objects from the Python 3.4+ pathlib module also expose these instance methods:

pathlib.Path.unlink() removes a file or symbolic link.
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We hope this article has helped you to resolve the problem. Apart from How to output a comma delimited list in jinja python template?, check other __del__-related topics.

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Angelo Wu

Massachussetts | 2023-03-26

Maybe there are another answers? What How to output a comma delimited list in jinja python template? exactly means?. Will use it in my bachelor thesis

Xu OConnell

Massachussetts | 2023-03-26

Thanks for explaining! I was stuck with How to output a comma delimited list in jinja python template? for some hours, finally got it done 🤗. I am just not quite sure it is the best method

Manuel Emmerson

Singapore | 2023-03-26

Simply put and clear. Thank you for sharing. How to output a comma delimited list in jinja python template? and other issues with code Python module was always my weak point 😁. I am just not quite sure it is the best method