[SOLVED] regex nagative assert does not work in Linux 'find' command with regextype being 'sed'

Issue

Say I want to match files where the second digit doesn't equal 0.

e.g.

./update110000.pt -> match
./update170000.pt -> match
./update100000.pt -> not match

so I wrote this

.*update[0-9](?!0)[0-9]{5}.pt

which passed the regex check on https://www.regexpal.com/

But when I use the find command on Linux like below it doesn't work anymore

find . -regextype sed -regex '.*update[0-9](?!0)[0-9]{5}.pt'

I also tried adding backslashes but didn't work either

find . -regextype sed -regex '.*update[0-9]\(?!0\)[0-9]{5}.pt'

I know that in this case I could simply use

find . -regextype sed -regex '.*update[0-9][1-9][0-9]{4}.pt'

which actually worked. But I still wonder how to properly use the (?!x) syntax in the find command.

Thanks!

Solution

If you define the regex flavor as sed, it is actually POSIX BRE that does not support lookarounds (you have a negative lookahead (?!0) in your pattern) and also follows some specific escaping rules:

find . -regextype sed -regex '\(.*/\)\{0,1\}update[0-9][1-9][0-9]\{4\}\.pt'

Details:

• \(.*/\)\{0,1\} - an optional sequence of any text and then /
• update - a fixed string
• [0-9] - a digit
• [1-9] - a non-zero digit (this is instead of the lookahead)
• [0-9]\{4\} - four digits** (note 4 here, not 5)
• \.pt - a .pt string.

If you use POSIX ERE, you can avoid overescaping:

find . -regextype posix-extended -regex '(.*/)?update[0-9][1-9][0-9]{4}\.pt'

Answered By - Wiktor Stribiżew

Answer Checked By - Marilyn (WPSolving Volunteer)