Using a variable in a expr regex command

Question

I am using the following bit of bash (sourced from here - I think)

bar=test_qux42_test
foo=(`expr ${bar} : '.*\(qux..\)'`)

The above returns qux42 successfully.

However, if I try the following it fails

baz=qux..
bar=test_qux42_test
foo=(`expr ${bar} : '.*\(${baz}\)'`)

I modify the command using a variable to customise the regex pattern and it fails. What am I doing wrong? How can I use a variable in the command?

Andrew Clark · Accepted Answer · 2013-07-24 16:56:57Z

2

Variables are not expanded inside of single quotes, try changing them to double quotes:

foo=(`expr ${bar} : ".*\(${baz}\)"`)

Or you can move the variable outside of the quotes:

foo=(`expr ${bar} : '.*\('${baz}'\)'`)

answered Jul 24, 2013 at 16:56

Andrew Clark

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Comments

chepner · Accepted Answer · 2013-07-24 16:59:59Z

1

There's no need to use expr for regex-matching in bash, which can perform it natively:

baz=qux..
bar=test_qux42_test
[[ $bar =~ .*\($baz\) ]]
foo=( "${BASH_REMATCH[1]" )

answered Jul 24, 2013 at 16:59

chepner

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