How to print matched regex pattern using awk? - regex

Question

Using awk, I need to find a word in a file that matches a regex pattern.
I only want to print the word matched with the pattern.
So if in the line, I have:
xxx yyy zzz
And pattern:
/yyy/
I want to only get:
yyy
EDIT:
thanks to kurumi i managed to write something like this:
awk '{
for(i=1; i<=NF; i++) {
tmp=match($i, /[0-9]..?.?[^A-Za-z0-9]/)
if(tmp) {
print $i
}
}
}' $1
and this is what i needed :) thanks a lot!

Answer 1

This is the very basic
awk '/pattern/{ print $0 }' file
ask awk to search for pattern using //, then print out the line, which by default is called a record, denoted by $0. At least read up the documentation.
If you only want to get print out the matched word.
awk '{for(i=1;i<=NF;i++){ if($i=="yyy"){print $i} } }' file

Answer 2

It sounds like you are trying to emulate GNU's grep -o behaviour. This will do that providing you only want the first match on each line:
awk 'match($0, /regex/) {
print substr($0, RSTART, RLENGTH)
}
' file
Here's an example, using GNU's awk implementation (gawk):
awk 'match($0, /a.t/) {
print substr($0, RSTART, RLENGTH)
}
' /usr/share/dict/words | head
act
act
act
act
aft
ant
apt
art
art
art
Read about match, substr, RSTART and RLENGTH in the awk manual.
After that you may wish to extend this to deal with multiple matches on the same line.

Answer 3

gawk can get the matching part of every line using this as action:
{ if (match($0,/your regexp/,m)) print m[0] }
match(string, regexp [, array])
If array is present, it is cleared,
and then the zeroth element of array is set to the entire portion of
string matched by regexp. If regexp contains parentheses, the
integer-indexed elements of array are set to contain the portion of
string matching the corresponding parenthesized subexpression.
http://www.gnu.org/software/gawk/manual/gawk.html#String-Functions

Answer 4

If Perl is an option, you can try this:
perl -lne 'print $1 if /(regex)/' file
To implement case-insensitive matching, add the i modifier
perl -lne 'print $1 if /(regex)/i' file
To print everything AFTER the match:
perl -lne 'if ($found){print} else{if (/regex(.*)/){print $1; $found++}}' textfile
To print the match and everything after the match:
perl -lne 'if ($found){print} else{if (/(regex.*)/){print $1; $found++}}' textfile

Answer 5

If you are only interested in the last line of input and you expect to find only one match (for example a part of the summary line of a shell command), you can also try this very compact code, adopted from How to print regexp matches using `awk`?:
$ echo "xxx yyy zzz" | awk '{match($0,"yyy",a)}END{print a[0]}'
yyy
Or the more complex version with a partial result:
$ echo "xxx=a yyy=b zzz=c" | awk '{match($0,"yyy=([^ ]+)",a)}END{print a[1]}'
b
Warning: the awk match() function with three arguments only exists in gawk, not in mawk
Here is another nice solution using a lookbehind regex in grep instead of awk. This solution has lower requirements to your installation:
$ echo "xxx=a yyy=b zzz=c" | grep -Po '(?<=yyy=)[^ ]+'
b

Answer 6

Off topic, this can be done using the grep also, just posting it here in case if anyone is looking for grep solution
echo 'xxx yyy zzze ' | grep -oE 'yyy'

Answer 7

Using sed can also be elegant in this situation. Example (replace line with matched group "yyy" from line):
$ cat testfile
xxx yyy zzz
yyy xxx zzz
$ cat testfile | sed -r 's#^.*(yyy).*$#\1#g'
yyy
yyy
Relevant manual page: https://www.gnu.org/software/sed/manual/sed.html#Back_002dreferences-and-Subexpressions

Answer 8

If you know what column the text/pattern you're looking for (e.g. "yyy") is in, you can just check that specific column to see if it matches, and print it.
For example, given a file with the following contents, (called asdf.txt)
xxx yyy zzz
to only print the second column if it matches the pattern "yyy", you could do something like this:
awk '$2 ~ /yyy/ {print $2}' asdf.txt
Note that this will also match basically any line where the second column has a "yyy" in it, like these:
xxx yyyz zzz
xxx zyyyz

Answer 9

echo "abc123def" | awk '
function MATCH(haystack, needle, ltrim, rtrim)
{
if(ltrim == 0 && !length(ltrim))
ltrim = 0;
if(rtrim == 0 && !length(rtrim))
rtrim = 0;
return substr(haystack, match(haystack, needle) + ltrim, RLENGTH - ltrim - rtrim);
}
{
print $0 " - " MATCH($0, "123"); # 123
print $0 " - " MATCH($0, "[0-9]*d", 0, 1); # 123
print $0 " - " MATCH($0, "1234"); # Nothing printed
}'