BASH: how to put variable inside regex?

Question

i'm trying to get working the following code:

searchfile="availables.txt"
read searchterm
grep_params="-i ^.*${searchterm}.*;.*$' $searchfile"
egrep $grep_params

which should echo all lines beginning with the $searchterm and followed by ";". But if the searchterm contains spaces it doesn't work (eg: "black eyed peas"), it gives me the following output:

egrep: eyed: No such file or directory
egrep: peas.*;.*$": No such file or directory
egrep: "availables.txt": No such file or directory

BASH FAQ entry #50: "I'm trying to put a command in a variable, but the complex cases always fail!" — Ignacio Vazquez-Abrams, Apr 19 '11 at 08:06

score 3 · Answer 1 · answered Apr 21 '11 at 00:23

3

Just Bash

searchfile="file"
read searchterm
shopt -s nocasematch
while read -r line
do
    case "$line" in
        *"$searchterm"*";"* ) echo "$line";;
    esac
done < "$searchfile"

answered Apr 21 '11 at 00:23

bash-o-logist

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Draco Ater · Answer 2 · 2011-04-19T08:27:54.453

1

You need to control word splitting here. That is done through arrays. See http://mywiki.wooledge.org/WordSplitting

searchfile="availables.txt"
read searchterm
grep_params=(-i "^.*${searchterm}.*;.*$" $searchfile)
egrep "${grep_params[@]}"

But don't use egrep - use grep -E instead, as the first is deprecated. But I would have changed your code like that:

searchfile="availables.txt"
read searchterm
grep_params="^.*${searchterm}.*;.*$"
grep -E -i "$grep_params" $searchfile

edited Apr 19 '11 at 08:27

answered Apr 19 '11 at 08:13

Draco Ater

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thanks, also: Is there any way to escape all spaces, brackets, etc.? So i can put in searchterm the string ".*" – Alessio Apr 19 '11 at 08:34
Didn't get you. What forbids you to enter `.*` when being asked for `searchterm`? – Draco Ater Apr 19 '11 at 08:41
if i enter ".*", it matches all the line – Alessio Apr 19 '11 at 08:46
Remove `.*$` in the grep_params, then it will match up to the ";" – Draco Ater Apr 19 '11 at 08:49
well, if i enter ".*", it should matches only lines that have "something.*something;something". Instead it matches all lines. I would have something that converts the string ".*" into "\.\*", in other words something that escape all the special character like .,\,/,*,$, etc – Alessio Apr 19 '11 at 08:54
1

http://stackoverflow.com/questions/2933474/escape-characters-contained-by-bash-variable-in-regex-pattern – Draco Ater Apr 19 '11 at 08:58

score 0 · Answer 3 · answered Apr 19 '11 at 08:12

The code you're looking for is something like:

searchfile="availables.txt"
read searchterm
regex="${searchterm}"'.*;'
egrep "${grep_params}" "${searchfile}"

Note that I simplified your regex (I hope I got it right!).

However, as Ignacio Vazquez-Abrar noted, this will fail in complex cases.

score 0 · Answer 4 · answered Apr 19 '11 at 08:16

0

As this is s regex, try to replace the " " with \s. This stands for a whitespace character.

"black\seyed\speas"

answered Apr 19 '11 at 08:16

stema

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This will also match other whitespace characters, which may not be what is intended. – Ignacio Vazquez-Abrams Apr 19 '11 at 08:17

score 0 · Answer 5 · answered Apr 19 '11 at 08:18

0

Alternative way is to pass the params through xargs:

searchfile="availables.txt"
read searchterm
grep_params="-i '^.*${searchterm}.*;.*$' $searchfile"
echo "$grep_params" | xargs egrep

answered Apr 19 '11 at 08:18

dogbane

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mmrtnt · Answer 6 · 2011-09-28T17:10:27.000

I got here by searching for "bash variable in a regex"

I solved it by changing the regex delimiters from "/" to "+"

Even though it has nothing to do with egrep, I'm adding my solution for other people who arrive from similar searches:

SYBASELOG="/opt/sybase/ASE-12_5/install/SYBASE.log"
MAILBODY="Some text and then the replacement placeholder:

   [MSGFILE]

   and some more text"

# Some proecessing...

MAILBODY=`echo "${MAILBODY}" | sed -e "s+\[MSGFILE\]+"${SYBASELOG}"+"`

And yes, I see now that it had little to do with bash and everything to do with the slashes in the log file variable. D'oh!

BASH: how to put variable inside regex?

6 Answers6