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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

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migrated from Apr 23 '11 at 22:04

This question came from our site for professional and enthusiast programmers.

up vote 1 down vote accepted

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'
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Use this syntax instead:


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}$/Test:rYYYYY/g' 
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It would be better to point that sed/awk/grep use POSIX regular expressions and why the additional -r is required. – mhitza Apr 23 '11 at 21:11
Also, the -r is not strictly necessary, but it changes the syntax somewhat if it is omitted: sed 's/Test:r[0-9]\{5\}/...' – William Pursell Apr 23 '11 at 23:50
Thanks guys, Any idea why sed -i 's/Test:r[0-9]\{5\}$/Test:rYYYYY/g' myfile.txt wouldnt work? Is the syntax wrong – smokinguns Apr 25 '11 at 4:31

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

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