CODEWITHSUNDEEP
pythonrestflask
ImpGod
ImpGod

Reputation: 1

How to make a PUT request with a full URL/link as a path parameter using Flask?

I need to add a link to a photo in my database using REST API. I have tried to add it as a string, but there is some error.

My PUT call:

http://127.0.0.1:3000/addphoto/3&http://www.clker.com/cliparts/8/m/S/S/c/x/number-3-coner-num-hi.png

and what I get:

127.0.0.1 - - [09/Apr/2022 09:10:51] "PUT /addphoto/3%26http://www.clker.com/cliparts/8/m/S/S/c/x/number-3-coner-num-hi.png HTTP/1.1" 404

Here is the code for the PUT route:

@app.route('/addphoto/<int:id>&<string:link1>', methods=['PUT'])
def addphoto(id,link1):
    connection = psycopg2.connect(db.data)
    cursor = connection.cursor()
    update_query = f"""UPDATE public."Items" SET photo = '{link1}' WHERE "Id" = {id}"""
    cursor.execute(update_query)
    connection.commit()
    return getitems()

Upvotes: 0

Views: 857

Answers (1)

danangjoyoo
danangjoyoo

Reputation: 360

It seems you want to pass it through path parameters. I think it would lead to high chances of errors since the route path matching algorithm has a bit of a complicated rule.

I suggest to pass it through query params like this

from flask import request

@app.route('/addphoto', methods=['PUT'])
def addphoto():
    id = request.args.get("id")
    link1 = request.args.get("link")
    connection = psycopg2.connect(db.data)
    cursor = connection.cursor()
    update_query = f"""UPDATE public."Items" SET photo = '{link1}' WHERE "Id" = {id}"""
    cursor.execute(update_query)
    connection.commit()
    return getitems()

Then you can hit it, such as this example request:

http://127.0.0.1:5000/addphoto?id=200&link=https://www.youtube.com/watch?v=ei7kbQhK1hA

Upvotes: 1

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