regex with 8 digits 2

[h3nd]

Hi guys,

I've just been wondering about to find 8 digit numbers

I've tried

if [[ $f == @([0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]) ]]

it's working fine, but looks abit stupid..
then, I've tried

if [[ $f == @(d\{8\}) ]] or
if [[ $f == @([0-9]\{8\}) ]]

Both above it's not working,
do you guys have any idea about this thanks?

 

[feherke]

Hi

As far as I know, there is no quantifier for given number of matches neither in [tt]bash[/tt] on [tt]ksh[/tt].

In such situations I use [tt]expr[/tt]. Note, that [tt]expr[/tt] is quite slow, so if you need it repeatedly, then better not use it.

if expr "$f" : '[[:digit:]]\{8\}$' > /dev/null
then
  [gray]# ...[/gray]
fi

Feherke.

http://rootshell.be/~feherke/

 

[h3nd]

Ok, thx feherke.

I think I better stick with my current pattern then..

Thanks man

 

[mrn]

what about

if [ ${#f} == 8 ]

Mike

Unix *is* user friendly. It's just selective about who its friends are.

 

[feherke]

Hi

Good point Mike.

Feherke.

http://rootshell.be/~feherke/