perl regex find and replace with variable from bash

Question

I have this regex: 'src=\d' which match all src attributes who start with a number in a file. I need to store it within a variable, cut src= out and write from there a new $string with \d concatenated to it: $string . $d. Is it possible to store only \d in a variable with a single command line? How to use cut and variable in a command line with perl? Is it possible?

perl -pi -w -e 's/src="\d+/src="http:\/\/website.com\/\d+/g’ file.tsv

Am I correct to assume that src=d is a typo src=\d?

Politank-Z
– Politank-Z

2015-05-31 00:03:03 +00:00
Commented May 31, 2015 at 0:03 — Politank-Z
– Politank-Z, Commented May 31, 2015 at 0:03

Toto · Accepted Answer · 2015-05-31 09:23:13Z

4

I'm not exactly sure what you mean, but I think you want something like this, where the () brackets store the number and the $1 replaces it back.

perl -pi -w -e 's/src="(\d+)/src="http:\/\/website.com\/$1/g’ file.tsv

And you can avoid the so-called 'leaning toothpick syndrome by selecting a different delimiter for the s/// operation like s{}{}

perl -pi -w -e 's{src="(\d+)}{src="http://website.com/$1}g’ file.tsv

edited May 31, 2015 at 9:23

Toto

91.7k63 gold badges97 silver badges135 bronze badges

answered May 30, 2015 at 23:56

Mort

3,5992 gold badges28 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

lizardhr Over a year ago

Yes, it is! I had tried this but without () brackets.. It only needs to be correct in the regex that is \d or \d+

user557597 Over a year ago

@lizardhr - just change the d+ to \d+

Collectives™ on Stack Overflow

perl regex find and replace with variable from bash

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related