Shell Programming and Scripting

Code:

echo 'String#1 and String#2' | egrep -o -m 1 'String#.{1}'
String#1
String#2

I'm trying to just match the first occurrence of 'String#' + 1 character. I thought the "-m 1" switch would do that for me. Instead I get both occurrences. Can somebody provide some insight?

Thanks!

Code:

echo 'String#1 and String#2' | awk '{for(i = 1; i <= NF; i++) {if($i ~ "^String#.$") {print $i; exit}}}'

The {1} seems redundant, if you leave it off it will also assume just one.

Code:

$ echo 'String#1 and String#2' | awk 'match($0, /String#./) { print substr($0, RSTART, RLENGTH); }'

String#1

$

1.

Code:

man egrep

sais:
so does not have an effect if your input is only 1 line.
Then
and the String#. matches twice, so both matches can be expected.
It looks like there is no way to have it stop at the first match within a line.
In this forum you can always expect an awk solution

Code:

echo 'String#1 and String#2' | awk 'BEGIN {RS=FS} $1~search {print $1; exit}' search="String#."

The exit means: exit after the first match.
BTW currently GNU awk and mawk and nawk do not know the {n} operator.
While {1} meaning "1 instance of the preceding character" can always be omitted, but {2} you need to write as ..

Just add a space at the end of the pattern:

Code:

echo 'String#1 and String#2' | egrep -o -m 1 'String#.{1} '
String#1

Thanks for the reply's! Script complete!