Regular expression in bash not working in conditional construct in Bash with operator '=~'

Question

The regular expression I have put into the conditional construct (with the =~ operator) would not return the value as I had expected, but when I assign them into two variables it worked. Wondering if I had done something wrong.

Version 1 (this one worked)

a=30
b='^[0-9]+$' #pattern looking for a number
[[ $a =~ $b ]]
echo $?

#result is 0, as expected

Version 2 (this one doesn't work but I thought it is identical)

[[ 30 =~ '^[0-9]+$' ]]
echo $?

#result is 1

Barmar · Accepted Answer · 2019-08-08 22:06:39Z

4

Don't quote the regular expression:

[[ 30 =~ ^[0-9]+$ ]]
echo $?

From the manual:

Any part of the pattern may be quoted to force the quoted portion to be matched as a string.

So if you quote the entire pattern, it's treated as a fixed string match rather than a regular expression.

answered Aug 8, 2019 at 22:06

Barmar

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3 Comments

Dennis Williamson Over a year ago

Also, I recommend putting the pattern in a variable and using the unquoted variable in the match expression: pattern='^[0-9]+$' and [[ 30 =~ $pattern ]] - this allows you to include characters such as whitespace in the pattern.

Barmar Over a year ago

@DennisWilliamson The question already shows how to do it using a variable, he was just wondering why it didn't work with a literal.

Dennis Williamson Over a year ago

Back to remedial reading for me! But it's useful to point out why a variable should be used and not a literal pattern.

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