String variable in a regular expression in bash?

Question

I have to write a regular expression in shell script to get a string inside another string, so that my variable string myString occurs in the regular expression string. How can I do this?

it would help if you tell more on what you want to do. what do you mean by "take myString"? do myString change, and what is constant in the context of the expression you want to match? — CharlesB
– CharlesB, Commented Sep 17, 2012 at 9:21
@CharlesB I edited my question. Could you please take a look ? — Larry
– Larry, Commented Sep 17, 2012 at 9:28
"myString is a constant string" ... unless I'm missing something here, you don't need regular expression if you're matching a constant string. Are you trying to extract everything inside the double quotes? Is the text outside the double quotes constant? — Shawn Chin
– Shawn Chin, Commented Sep 17, 2012 at 9:29

Shawn Chin · Accepted Answer · 2012-09-17 09:33:48Z

2

If you want to extract text within the double quotes, and assuming there's only one set of double quotes, one way to do that is:

[me@home]$ echo $A
to get "myString" in regular expression
[me@home]$ echo $A | sed -n 's/.*"\(.*\)".*/\1/p'
myString

Of course, if there's only one set of quotes you can also do without sed/regex:

[me@home]$ echo $A | cut -d'"' -f2
myString

answered Sep 17, 2012 at 9:33

Shawn Chin

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Comments

Thor · Accepted Answer · 2012-09-17 10:42:19Z

1

If you know there will only be one set of double quotes, you could use shell parameter expansion like this:

zsh> s='to get "myString" in regular expression'
zsh> echo ${${s#*\"}%\"*}
mystring

bash doesn't support multilevel expansion, so the expansion needs to be applied sequentially:

bash> s='to get "myString" in regular expression'
bash> s=${s#*\"}
bash> s=${s%\"*}
bash> echo $s
mystring

answered Sep 17, 2012 at 10:42

Thor

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Comments

Vijay · Accepted Answer · 2012-09-17 10:34:30Z

0

>echo 'hi "there" ' | perl -pe 's/.*(["].*["])/\1/g'
"there"

answered Sep 17, 2012 at 10:34

Vijay

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akozin · Accepted Answer · 2012-09-17 10:55:41Z

0

You can also use 'awk':

echo 'this is string with "substring" here' | awk '/"substring"/ {print}'

# awk '/"substring"/ {print}' means to print string, which contains regexp "this"

answered Sep 17, 2012 at 10:55

akozin

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Douglas Santos · Accepted Answer · 2016-09-27 19:59:13Z

0

In Bash, you can use the =~ operator in a [[ ... ]] conditional construct, together with the BASH_REMATCH variable.

Example of use:

TEXT='hello "world", how are you?'
if [[ $TEXT =~ \"(.*)\" ]]; then
    echo "found ${BASH_REMATCH[1]} between double quotes."
else
    echo "nothing found between double quotes."
fi

answered Sep 27, 2016 at 19:59

Douglas Santos

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Comments

Bitwise · Accepted Answer · 2012-09-17 09:27:04Z

-1

grep is the most common tool for finding regular expressions in the shell.

answered Sep 17, 2012 at 9:27

Bitwise

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1 Comment

Larry Over a year ago

yeap, I know I can use grep like this : grep -P '.*".*".*' -o [FILE] however, I wanna to get string inside double quotes. @Bitwise

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