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How do I get the different parts of a Flask request's url?

Tags:

python

url

flask

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How do I find the base URL of a Flask?

You can use the base_url method on flask's request function. 127.0. 0.1 - - [28/Mar/2017 10:44:53] "GET /flights HTTP/1.1" 200 -... if want to return whole thing , then what have i to do ? @RaviBhushan I'm not aware of any method to capture the standard output of the flask server.

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The url_for() function is used to build a URL to the specific function dynamically. The first argument is the name of the specified function, and then we can pass any number of keyword argument corresponding to the variable part of the URL.

How do you get the request object in Flask?

To access the incoming data in Flask, you have to use the request object. The request object holds all incoming data from the request, which includes the mimetype, referrer, IP address, raw data, HTTP method, and headers, among other things.


You can examine the url through several Request fields:

Imagine your application is listening on the following application root:

http://www.example.com/myapplication

And a user requests the following URI:

http://www.example.com/myapplication/foo/page.html?x=y

In this case the values of the above mentioned attributes would be the following:

    path             /foo/page.html
    full_path        /foo/page.html?x=y
    script_root      /myapplication
    base_url         http://www.example.com/myapplication/foo/page.html
    url              http://www.example.com/myapplication/foo/page.html?x=y
    url_root         http://www.example.com/myapplication/

You can easily extract the host part with the appropriate splits.


another example:

request:

curl -XGET http://127.0.0.1:5000/alert/dingding/test?x=y

then:

request.method:              GET
request.url:                 http://127.0.0.1:5000/alert/dingding/test?x=y
request.base_url:            http://127.0.0.1:5000/alert/dingding/test
request.url_charset:         utf-8
request.url_root:            http://127.0.0.1:5000/
str(request.url_rule):       /alert/dingding/test
request.host_url:            http://127.0.0.1:5000/
request.host:                127.0.0.1:5000
request.script_root:
request.path:                /alert/dingding/test
request.full_path:           /alert/dingding/test?x=y

request.args:                ImmutableMultiDict([('x', 'y')])
request.args.get('x'):       y

you should try:

request.url 

It suppose to work always, even on localhost (just did it).


If you are using Python, I would suggest by exploring the request object:

dir(request)

Since the object support the method dict:

request.__dict__

It can be printed or saved. I use it to log 404 codes in Flask:

@app.errorhandler(404)
def not_found(e):
    with open("./404.csv", "a") as f:
        f.write(f'{datetime.datetime.now()},{request.__dict__}\n')
    return send_file('static/images/Darknet-404-Page-Concept.png', mimetype='image/png')