3

I have a file in which each line is like this

"372"^""^"2015-09-03 06:59:44.475"^"NEW"^"N/A"^""^0^"105592"^"https://example-url.com"^"example-domain < MEN'S ULTRA < UltraSeriesViewAll (18)"^"New"^"MERCHANT_PROVIDED"

I want to extract the urls in the file -- https://example-url.com

I tried these regex using sed command -- sed -n '/"^"http/,/"^"/p'

But it didn't solve my problem.

4

You could use this

sed -n 's!^.*\^"\(http[^^]*\)"^.*!\1!p'

The potential gotcha for a beginner to REs is that ^ is an indicator for start of line, so you have to ensure you escape it \^ if you want a literal up arrow at the start of your RE.

The RE pattern match can be explained as follows

  • ^.*\^" -- Match from start of line until we see the last possible up-arrow double-quote ^" that satisfies the rest of the pattern
  • \( -- Start a substitution block that can be substituted as \1
  • http[^^]* -- Match http followed by as many characters that are not ^ as possible
  • \) -- End the substitution block "^.* -- Match double-quote and up-arrow "^, then as much as possible (until end of line)

This entire match is replaced by \1, which is the pattern block starting http

3

Try this:

echo "372"^""^"2015-09-03 06:59:44.475"^"NEW"^"N/A"^""^0^"105592"^"https://example-url.com"^"example-domain < MEN'S ULTRA < UltraSeriesViewAll (18)"^"New"^"MERCHANT_PROVIDED" | cut -f9 -d^
  • I think you meant to wrap your echo command in single quotes '. Otherwise you lose the double quotes when you echo. – Centimane Sep 4 '15 at 12:21
  • Given statement is also equipped with single quote '. Hence I avoided that. – SHW Sep 4 '15 at 12:23
  • But without the single quote fields like f 10 become arrays. It doesn't impact the url itself really, but it's easy enough to echo the result of this to get rid of the double quotes if needed. – Centimane Sep 4 '15 at 12:29
  • The cut is useful but doesn't remove the double quotes surrounding the extracted field. tr -d '"' maybe? – roaima Sep 4 '15 at 13:20
3

If your version of grep supports PCRE mode, you could try

grep -Po '(?<="\^")http.+?(?="\^")'
0

If your URLs always start with http and end with a quote, you can just search for the string http and everything up to the next quote:

  1. grep

    $ grep -o 'https*://[^"]*' file
    https://example-url.com
    
  2. sed

    $ sed -n 's#.*\(https*://[^"]*\).*#\1#;p' file
    https://example-url.com
    
  3. Perl

    $ perl -ne 's#.*(https*://[^"]*).*#\1# && print' file
    https://example-url.com
    
  4. With a slightly different approach, you can use awk as well. Just use -F to set the field delimiter to " and print any field starting with hhtp:

    $ awk -F\" '{for(i=1;i<NF;i++){if($i~/^http/){print $i}}}' file
    https://example-url.com
    

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