Flask send variable data to python through url_for()

Question

I want to send a variable value to python from flask html/js through url_for().

Python Code :

@app.route('/video_feed/<device>')
def video_feed(device):
    # return the response generated along with the specific media
    # type (mime type)
    print(device)
    try:
        return Response(gen(int(device)),mimetype = "multipart/x-mixed-replace; boundary=frame")
    except Exception as e:
        return Response(gen(device),mimetype = "multipart/x-mixed-replace; boundary=frame")

Desired in flask html file:

data.value = 0;
image.src = "{{ url_for('video_feed', device=data.value)}}";

But it does not work, I get the error:

jinja2.exceptions.UndefinedError: 'data' is undefined

This does work, however I can't use the variable data.value in url_for() :

image.src = "{{ url_for('video_feed', device=0)}}";

I have tried several different things like:

image.src = "{{ url_for('video_feed', device=${data.value})}}";
image.src = "{{ url_for('video_feed', device=$data.value)}}";
image.src = "{{ url_for('video_feed', device=%s)}}", data.value;

But nothing seems to work. My javascript is a bit rusty.

Any help would be appreciated!

Cheers!

Dave W. Smith · Accepted Answer · 2020-02-27 03:30:03Z

1

url_for is expanded on the server side, before the page gets sent to the browser and well before JavaScript runs.

If you know the number (for device/data.value) at page generation time, pass it in to the template via render_template().

If, however, you don't know the value until after the page is rendered, you're going to need to construct the img element from JavaScript.

answered Feb 27, 2020 at 3:30

Dave W. Smith

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3 Comments

Grebtsew Over a year ago

Thanks for your reply. I do not know the value until after the page is rendered. This is my scenario. The Flask Website receive messeges via SocketIO in JSON format. The "data" variable above is that JSON object. The field "value" contains an address for a captureable device. For each incoming JSON object I want to create a new <img> with src=url_for(...). As i mention this works when I'm not using variables. I agree that it's because of when the page is rendered. Can I do this some other way? I have a couple of ideas like, ajax? or proxy-server or restructure to solve this on the python side.

Grebtsew Over a year ago

Oh, now I see what you mean the src generated by url_for() are replicable in js.

Grebtsew Over a year ago

So there were really no need for me to use url_for(). Should just use image.src = "/video_feed/"+data.value;

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