Why sed is not honouring the regex replace request-CodePudding

I have a grep result like this

Nov-06-22 00:01:16 id-03674-09704 
Nov-06-22 00:01:16 id-03642-04246

and I need to remove from result the id-.....-..... part.

I am using this:

grep file.txt | sed 's/ id-(\d){5}-(\d){5}/ /g'

but it returns this

Nov-06-22 00:01:16 id-03674-09704 
Nov-06-22 00:01:16 id-03642-04246

I checked the regex id-(\d){5}-(\d){5} and it should be ok.

Why sed is not replacing the grep result?

CodePudding user response：

Since you are using POSIX BRE regex flavor with sed, \d are not recognized as digit matching construct, and {5} are treated as literal {5} strings, not interval quantifiers.

You need to replace \d with [0-9] and use the -E option to enable POSIX ERE syntax (or escape the interval quantifier braces):

sed -E 's/ id-[0-9]{5}-[0-9]{5}//' file
sed 's/ id-[0-9]\{5\}-[0-9]\{5\}//' file

See the online demo:

#!/bin/bash
s='Nov-06-22 00:01:16 id-03674-09704 
Nov-06-22 00:01:16 id-03642-04246'
sed -E 's/ id-[0-9]{5}-[0-9]{5}//' <<< "$s"

Output:

Nov-06-22 00:01:16 
Nov-06-22 00:01:16

Also, consider just removing last column with awk, or even with cut:

awk 'NF{NF-=1};1' file.txt
cut -d' ' -f1,2 < file.txt

CodePudding user response：

You can simply do it using cut by specifying the delimiter which I suppose is space in your case. You first need to read your file and do the grep then use cut command. Suppose you already have a file with the selected lines then you can use cut as below:

cat file.txt | cut -d' ' -f1,2

output:

Nov-06-22 00:01:16
Nov-06-22 00:01:16