Hi I want to print the line until pattern is matched.
I am using below code:
sed -n '1,/pattern / p' file
It is working fine for me , but its not working for exact match.
sed -n '1,/^LAC$/ p' file
Input:
LACC FEGHRA 0
LACC FACAF 0
LACC DARA 0
LACC TALAC 0
LAC ILACTC 0
LACC Kerzner 0
LAAS TOTAL 0
LACC LAC 0
LACC IRSI 0
LACC Asocianes CTO 0
LACC Kerzner 0
Expected Output:
LACC FEGHRA 0
LACC FACAF 0
LACC DARA 0
LACC TALAC 0
LAC ILACTC 0
That is because $ would match end of line and not end of string. Depending on the exact nature of the input you could match for space/tab instead, or just not a letter:
sed -n '1,/^LAC[^a-zA-Z]/p' file
And if the Input is like below:
LACC FEGHRA 0
LACC FACAF 0
LACC DARA 0
LACC TALAC 0
LACC ILACTC 0
LACC Kerzner 0
LAAS TOTAL 0
LACC LAC 0
LACC IRSI 0
LACC Asocianes CTO 0
LACC Kerzner 0
Then Can I achive the below output:
LACC FEGHRA 0
LACC FACAF 0
LACC DARA 0
LACC TALAC 0
LACC ILACTC 0
LACC Kerzner 0
LAAS TOTAL 0
LACC LAC 0
Thank you
Although this
sed -n '1,/ LAC/p' file
will work with the sample given, try this
sed -rn '1,/^[^ ]+[ ]+LAC[ ]+/p' file
to be on the safe side (within the square brackets, there's a space and a <TAB> char). If your sed doesn't support the -r (or -E) option, replace the + with \+ .
Hi,
with gnu sed, we can use begin mark and end mark word:
sed -n '1,/\<LAC\>/p' file
Regards.
sed -n '1,/\<LAC\>/p' file
This doesn't work for me.
Only the below code is working.
sed -n '1,/ LAC/p' file
Ok, then a posix solution:
sed -n '1,/[^[:alnum:]_]LAC[^[:alnum:]_]/p' file
PS: posix definiton of word class is [A-Za-z0-9_]
Regards.
With awk:
awk '{print; if ($1=="LAC") exit}' file
awk '{print; if ($2=="LAC") exit}' file
awk '{print; if ($1=="LAC" || $2=="LAC") exit}' file
But there is one catch in the case of sed. If the match is on the first line, then it will not work.. To handle this case you need a GNU extension (GNU sed):
gsed -n '0,/\<LAC\>/p' file
awk does not have this problem and can also use a range expression and also can have an exact field match as noted by MadeInGermany, so you do not need to delimit the search:
awk 'NR==1, $1=="LAC"||$2=="LAC"' file
or
awk -v s=LAC 'NR==1, $1==s||$2==s' file
MadeinGermany's approach with the exit statement would be more efficient.
--
This exit approach would also work with regular sed and this does not have the match-on-line-1 catch:
sed '/^LAC$/q; /^LAC[[:blank:]]/q; /[[:blank:]]LAC[[:blank:]]/q' file