regex getting url link

Question

I am trying to extract the link I am getting from a curl command. Curl command throws back of type string.

{"success":true,"key":"Syv77d","link":"https://file.io/Syv77d","expiry":"14 days"}

In my below code this gets https://file.io/Syv77d","expiry":"14 days"}
link = re.search('https://.*$',fileIO)

What I wanted was just https://file.io/Syv77d

The link would vary so i would need the url without the double-qoutes. I think I am missing something in my regex.

Use a JSON parser? Just get `json.loads(x)["link"]` where `x` is the JSON — ctwheels, Feb 22 '18 at 15:45
It appears the response is in [JSON](https://json.org/) format. You may consider using Python's `json` module instead of manually parsing. — Cong Ma, Feb 22 '18 at 15:45

score 1 · Accepted Answer · answered Feb 22 '18 at 15:45

1

Convert the string object to a JSON object.

Ex:

import json
jData = json.loads('{"success":true,"key":"Syv77d","link":"https://file.io/Syv77d","expiry":"14 days"}')
jData["link"]

answered Feb 22 '18 at 15:45

Rakesh

yes i have done exactly like this. thanks as well – Dee Feb 22 '18 at 15:49
You are welcome. Please accept ans if it solved your problem. Thanks – Rakesh Feb 22 '18 at 16:02

1 Answers1