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pythoncssflaskreturn

Python Flask "return" is not applying css style

I'm new in python / flask, so I'm probably not solving the issue the way it should be solved... Considering this, the question is:

I have a flask def(), which gets a POST from an HTML, and returns the text "File successfully processed".

What I want is for the "File successfully processed" to have the style in my styles.css file.

PS.: I already have an index.html which works great with the styles.css. The code in index.html is:

  <head>
  <link rel= "stylesheet" type= "text/css" href= "{{ url_for('static',filename='css/styles.css') }}">
  </head>

So, how do I apply this styles.css to the return "File successfully processed"?

@app.route("/run_ctb", methods=["GET", "POST"])
def web_run_ctb():
if request.method == "POST":
    return "<p>File successfully processed<p>"
Solution

  • That styles sheet is currently only applying to your index.html template. You should create a template for your "/run_ctb" page, for example called run_ctb.html which would look like:

    <head>
    <link rel= "stylesheet" type= "text/css" href= "{{ 
    url_for('static',filename='css/styles.css') }}">
</head>

<body>
    <p>File successfully processed<p>
</body>

    Then you can just have your route function render the template:

        @app.route("/run_ctb", methods=["GET", "POST"])
    def web_run_ctb():
        if request.method == "POST":
            return render_template('run_ctb.html')