Why doesn't this simple bash regex return true?

Question

If I do [[ "0" =~ "^[0-9]+$" ]] && echo hello at a terminal I would expect to see the word "hello"

However, nothing gets printed. What am I doing wrong?

[[ "0" =~ ^[0-9]+$ ]]

Avinash Raj
– Avinash Raj

2015-05-21 08:20:34 +00:00
Commented May 21, 2015 at 8:20 — Avinash Raj
– Avinash Raj, Commented May 21, 2015 at 8:20
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Tom Fenech
– Tom Fenech

2015-05-21 09:04:45 +00:00
Commented May 21, 2015 at 9:04 — Tom Fenech
– Tom Fenech, Commented May 21, 2015 at 9:04

Avinash Raj · Accepted Answer · 2015-05-21 08:21:01Z

5

You need to remove the double quotes present in your regex. ie, don't enclose your regex pattern within double quotes.

[[ "0" =~ ^[0-9]+$ ]]

answered May 21, 2015 at 8:21

Avinash Raj

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Comments

Maroun · Accepted Answer · 2015-05-21 08:21:17Z

3

It should be:

[[ "0" =~ ^[0-9]+$ ]] && echo hello

Note that the second part is not surrounded with double quotes, otherwise it'll be treated as the string "^[0-9]+$" and not a regex. To confirm that, try:

[[ "^[0-9]+$" =~ "^[0-9]+$" ]] && echo hello

answered May 21, 2015 at 8:21

Maroun

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1 Comment

chepner Over a year ago

Slight technicality: the right hand side is still treated as a regular expression ([[ foo =~ "fo" ]] would still succeed, for example). The difference is that any metacharacters inside the regex will be treated literally.

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