Do not replace the second occurrence of pattern in a same word in the file with sed command

Question

echo "abcdef12cade 12345678 abcdefba12345678" | sed -E "s/[0-9a-fA-F]{8}/00000000/g" 

the above outputs the following

00000000cade 00000000 0000000000000000

it is replacing the pattern for second occurrence in a same word. I don't want to replace if there is a second occurrence.

expected ouput

00000000cade 00000000 abcdefba12345678

Answer 1

If you only want to replace the first occurrence of a match, don't use the g suffix to the command:

$ echo 'aa' | sed 's/a/b/g'
bb
$ echo 'aa' | sed 's/a/b/'
ba

The g option stands for 'global', which is explicitly telling sed to replace all matches and not just the first (which is the default behavior).

Answer 2

It seems like you're looking for words that are between 8 and 15 chars, and replace the first 8 hex digits:

sed -E 's/\<[[:xdigit:]]{8}([[:xdigit:]]{0,7})\>/00000000\1/g' <<END
abcdef12cade 12345678 abcdefba12345678 12345 123456789
END

00000000cade 00000000 abcdefba12345678 12345 000000009

Where, \< and \> are word boundaries, and [:xdigit:] matches a hex digit.

Answer 3

The g at the end make sed repeat the substitution as many times as possible on the line. You only want to do it twice.

Let's do that with GNU awk:

echo 'abcdef12cade 12345678 abcdefba12345678' |
awk '{ sub("[0-9a-fA-F]{8}", "00000000", $1)
       sub("[0-9a-fA-F]{8}", "00000000", $2)
       print }'

This performs the substitution on the two first whitespace-delimited fields only, and then prints the resulting line.