How to check if a string only contains A-Z, a-z and 0-9?

Question

What is the best way to validate a string with a pattern? I would use PCRE but I don't know if it is embedded in each shell and how to use it.

For example, how could I validate that variable only contains A-Z, a-Z and 0-9 and does not contain spaces, ', ", ... ?

Solution 1

$ [[ "foo" =~ ^[A-Za-z0-9]*$ ]] ; echo $?
0
$ [[ "foo " =~ ^[A-Za-z0-9]*$ ]] ; echo $?
1

Solution 2

if `echo $VARIABLE | egrep '[^A-Za-z0-9]'`; then echo VARIABLE IS BAD; fi

A pure shell option

case "$VARAIBLE" in *[^A-Za-z0-9]*) echo VARIABLE IS BAD;; esac

Solution 3

if [[ "$VARIABLE" =~ ^[[:alnum:]]*$ ]]; then do something; fi;

useful resources: http://bashshell.net/regular-expressions/ , http://www.gnu.org/software/bash/manual/bashref.html

Comments

[Chris Eberle]

Generally in bash you'll want to rely pretty heavily on grep or awk to do any pattern matching.

[Gordon Davisson]

Several corrections: there cannot be spaces between the outer double-brackets, the [:alnum:] needs another set of brackets, and should be followed by * to let it match more than one character. Fixed version: if [[ "$VARIABLE" =~ ^[[:alnum:]]*$ ]]; then do something; fi

[Martin]

Thank you for correcting the syntax. Updated as to not confuse.

[Ignacio Vazquez-Abrams]

Be careful with this one. $ [[ "一" =~ ^[[:alnum:]]*$ ]] ; echo $? 0

[Gordon Davisson]

@Ignacio: that prints 1 for me (as expected). Is 一 in the alnum character class in some other locale (I'm using LANG=en_US.UTF-8)?

[Ignacio Vazquez-Abrams]

If your bash uses Unicode classes then it will be true, as 一 is "1" in Japanese.