SED: Search a alphanumeric string and replace it

Question

Guys, I need to search and replace an alphanumeric string using SED.

Search String: Test:rXXXXX, where XXXXX will always be a 5 digit number

Replace String: Test:rYYYYY, where YYYYY will always be a 5 digit number

I have come up with the following :

echo 'Test:r12345' | sed 's/Test:r[0-9][0-9][0-9][0-9][0-9]$/Test:rYYYYY/g'

This works currently.

Is there a better way to achieve this? I don't want to use '[0-9]' 5 times in the expression

Answer 1

Just another one with sed with a small modification to match the "Test:r" subexpression making the replacement a bit more terse

echo 'Test:r12345' | sed 's/\(Test:r\)[0-9]\{5\}/\1YYYYY/g'

Answer 2

Use this syntax instead:

[0-9]{5}

This requires you to pass '-r' to sed, so the new example will be:

echo 'Test:r12345' | sed -r 's/Test:r[0-9]{5}$/Test:rYYYYY/g' 

Answer 3

well, based on your sample, you can just replace the 2nd field using awk

echo 'Test:r12345' | awk -F":" '{$2="rYYYYY"}1' OFS=":"

There is no need to create complex regular expression. KISS