Clean Urls with regular expression

Question

i have Data in thousand lines like

http://xxxx.com/xxx-xxx-xxx-xxxx/ 60% 2 Weekly 2014-01-01 00:00

want to remove everything after / in every url

(output should be in clean url like below)

http://xxxx.com/xxx-xxx-xxx-xxxx/

Thanks

Answer 1

Use the replace menu by pressing Ctrl+H, and make sure regular expressions are enabled. Then,

Find (^.*\/).* and Replace $1: https://regex101.com/r/lJ4lF9/12

Alternatively, Find (?m)(^.*\/).* and Replace $1: https://regex101.com/r/lJ4lF9/13

Explanation:

Within a capture group, Find the start of the string (^) followed by anything any number of times (.*) until the last "/", then anything any number of times. Replace with the captured group by referencing it as $1.

(?m)

Answer 2

One way to do that is use linux command line:

cat file.txt |cut -f1 -d" "

If you are interested in regex then this will match the url in a single line:

[^\ ]+