Using regex quantifier with find [duplicate]

Question

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I've got a set of files in a directory, all of the format test[0-9][0-9].txt. If I run

find . -regex ".*/(test)[0-9][0-9]\.txt"

then all the files are shown, but if I run

find . -regex ".*/(test)[0-9]{2}\.txt"

then none are shown. What am I doing wrong?!

I've had a search through previous similar questions to this, but can't find a particularly relevant answer

The first example uses normal regexp syntax, where the second uses extended syntax. Your version of find obviously expects the former by default. You don't say what system you're on - your find might have a switch to turn on extended regexp parsing. On FreeBSD, for example, it's -E — D_Bye, Nov 23 '15 at 12:27
@D_Bye ah, that's a good point. CentOS6. I've tried -regextype posix-extended and that doesn't work either — ChrisW, Nov 23 '15 at 12:33

the_velour_fog · Accepted Answer · 2015-11-23 13:28:44Z

If you are using GNU find, this should work for you

 find . -regextype sed -regex "./test[0-9]\{2\}.txt"

-regextype sed - use basic posix regular expression (just because thats what Im familiar with)
./ - necessary because find considers all file paths on a relative search to begin with this pattern
[0-9]\{2\} - 2 instances of the [0-9] digit character class with the quantifier \{...\} brackets escaped - per basic posix regular expressions

marked as duplicate by don_crissti, slm♦ Nov 23 '15 at 16:12