I am in shell and I have this string: 12 BBQ ,45 rofl, 89 lol
Using the regexp: \d+ (?=rofl), I want 45 as a result.
Is it correct to use regex to extract data from a string? The best I have done is to highlight the value in some of the online regex editor. Most of the time it remove the value from my string.
I am investigating expr, but all I get is syntax errors.
How can I manage to extract 45 in a shell script?
You can do this with GNU grep's perl mode:
echo "12 BBQ ,45 rofl, 89 lol" | grep -P '\d+ (?=rofl)' -o
echo "12 BBQ ,45 rofl, 89 lol" | grep --perl-regexp '\d+ (?=rofl)' --only-matching
-P and --perl-regexp mean Perl-style regular expression.
-o and --only-matching mean to output only the matching text.
docker port c62c1c7b9efb | grep -P '(\d+)$' -o
-E instead of -P
-E instead of -P, but I believe that won't work, since -P in GNU Grep is for Perl mode, while -E in (most?) greps is extended mode, which is quite different. I just tested a different version's -E option and it output nothing instead of the expected 45.Yes regex can certainly be used to extract part of a string. Unfortunately different flavours of *nix and different tools use slightly different Regex variants.
This sed command should work on most flavours (Tested on OS/X and Redhat)
echo '12 BBQ ,45 rofl, 89 lol' | sed 's/^.*,\([0-9][0-9]*\).*$/\1/g'
.*-P grep option
/bin/sh -c "command". Works nicely!
It seems that you are asking multiple things. To answer them:
You can extract the numbers by catching them in capturing parentheses:
.*(\d+) rofl.*
and using $1 to get the string out (.* is for "the rest before and after on the same line)
With sed as example, the idea becomes this to replace all strings in a file with only the matching number:
sed -e 's/.*(\d+) rofl.*/$1/g' inputFileName > outputFileName
or:
echo "12 BBQ ,45 rofl, 89 lol" | sed -e 's/.*(\d+) rofl.*/$1/g'
.* in your example. You only need those on edges if your regex is anchored. Unanchored, it will already match anywhere inside the string.
.*, it's a simple way to match everything and replace by what's in the matching parentheses. Without them, the rest of the string remains intact, which is not what was asked (iiuc). Or did I miss something perhaps?sed for this. Carry on.$1 in crazinessseds that come with Ubuntu 14.04, Ubuntu 18.04 and FreeBSD 11 (both because of the non-portable \d and that fact that $1 is treated literally in this context)you can use the shell(bash for example)
$ string="12 BBQ ,45 rofl, 89 lol"
$ echo ${string% rofl*}
12 BBQ ,45
$ string=${string% rofl*}
$ echo ${string##*,}
45
You can certainly extract that part of a string and that's a great way to parse out data. Regular expression syntax varies a lot so you need to reference the help file for the regex you're using. You might try a regular expression like:
[0-9]+ *[a-zA-Z]+,([0-9]+) *[a-zA-Z]+,[0-9]+ *[a-zA-Z]+
If your regex program can do string replacement then replace the entire string with the result you want and you can easily use that result.
You didn't mention if you're using bash or some other shell. That would help get better answers when asking for help.
You can use rextract to extract using a regular expression and reformat the result.
Example:
[$] echo "12 BBQ ,45 rofl, 89 lol" | ./rextract '[,]([\d]+) rofl' '${1}'
45
