grep pattern and the content after that, and remove others

Question

I'm having a problem to grep specific keyword and its content. This is sample file, the actual file is larger than this.

Example

user@linux:~$ cat url.txt 
abcrandomtextdef another random text blablabla
another random iwantthis text abcrandomtextdef url=https://www.google.com ghirandomtextjkl
ghirandomtextjkl another random text yadayada
wxyz iwantthis abcdef url=yahoo.com yaday
user@linux:~$ 

Desired output would be like this.

iwantthis url=https://www.google.com
iwantthis url=yahoo.com

These are my attempts to get that output, but as you can see it didn't really work.

user@linux:~$ grep url= url.txt | cut -d ' ' -f3,6
iwantthis url=https://www.google.com
abcdef
user@linux:~$ 

Answer 1

sed would seem to be the right task for this:

% sed -n 's/.* \(iwantthis\) .* \(url=[^ ]*\) .*/\1 \2/p' url.txt
iwantthis url=https://www.google.com
iwantthis url=yahoo.com

How this works:

-n -- only print lines that match with a "p" command

s/.../p -- search and replace, printing lines that match

.* \(iwantthis\) .* \(url=[^ ]*\) .* -- This will look for the word "iwantthis" surrounded by spaces and remember it, and also look for "url=" followed by non-spaces, and rememeber that. The .* at each end mean that stuff before "iwantthis" and stuff after the URL are discarded.

/\1 \2 -- Replace it with the two remembered words