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I'm facing a problem and trying to find a solution that works in sh. If I could use bash this code would work:

ls /a|grep ^[0-9]

Unfortunately this is not the case with sh, and yes I need to use sh. :) Running in sh i get

ls /a |grep ^[0-9]
[0-9]: not found
Usage: grep -hblcnsviw pattern file . . .

If I remove the ^ the code works but I need only the files that start with numbers and not the ones that contains numbers. Example, I need the files that are like:


from the grep man page I should be able to use ^.

For the time being my implementation was done by using:

ls /a|grep -v [a-z]|grep -v [A-Z]

As this will remove all the files that contains chars, but still if a file is .123.33 it will show up.

share|improve this question
Can you use egrep instead of grep? – slm Jun 5 '13 at 12:48
tip: grep -v [a-z]|grep -v [A-Z] == grep -v "[a-z]\|[A-Z]" – rush Jun 5 '13 at 12:59
it wouldn't work well, I tried with egrep and no luck, but the solution provided by slm works like a charm :) – BitsOfNix Jun 5 '13 at 13:03
@rush, that syntax is not portable and will not work in the OP's Solaris. grep -v '[a-zA-Z]' or grep -ve '[a-z]' -e '[A-Z]' or grep -vE '[a-z]|[A-Z]' are standard though (though the behaviour is locale dependant). – Stéphane Chazelas Jun 5 '13 at 13:44
up vote 3 down vote accepted

Something like this should work for you:

ls /a | egrep "^[0-9]"

Per @Anthons feedback egrep is deprecated, so you can use -E as a switch to the normal grep command instead:

ls /a | grep -E "^[0-9]"

Per @Stephane's feedback the extended regular expressions (ERE's) aren't even necessary in this situation. Really what matters is the quoting of the ^[0-9] to protect it from being interpreted by whatever version of /bin/sh you're using that's having the issue. So something like this would be the simplest fix to your issue:

ls /a | grep "^[0-9]"


ls /a | grep '^[0-9]'
share|improve this answer
egrep is deprecated (according to the man page), should recommend the use of grep -E – Anthon Jun 5 '13 at 13:01
so I guess that bash interprets the ^ while when using sh it is up to grep, correct? and thanks a lot. – BitsOfNix Jun 5 '13 at 13:02
Yes, bash is interpreting those instead of grep. I was surprised that you could pass those arguments bare like that w/o double quotes. I've only ever double quoted them to grep. You could've also put single quotes around them too. – slm Jun 5 '13 at 13:04
You don't need extended REs here. That syntax works as well with BREs as with EREs so the e or -E are superfluous. – Stéphane Chazelas Jun 5 '13 at 13:47

By sh, I think you're refering to the Bourne shell which was the shell of most Unix systems before the mid 90s and was /bin/sh on Solaris prior to Solaris 11.

On Solaris 10 and older, don't use /bin/sh. That shell is from another era. Use /usr/xpg4/bin/sh instead.

In the Bourne shell ^ is an alias for | for compatibility with its predecessor the Thomson shell. So your command is like:

ls /a|grep |[0-9]

And the Bourne shell reports that it can't find a command called [0-9] and grep complains about not getting any argument.

Even if using a standard sh as opposed to the Bourne shell, I would recommend that you quote ^. For instance ^ is a globbing operator in zsh when the extendedglob option is enabled.

In any case, if not ^, you have to quote [0-9] since those are globbing operators. [0-9] would be expanded by the shell to the list of files in the current directory whose name is a single digit.


ls /a | grep '^[0-9]'

Incidentally, in the Bourne shell

ls /a ^ grep '^[0-9]'

would also work.

share|improve this answer
Thanks for the lesson 8-). – slm Jun 5 '13 at 14:07
Thanks a lot for the explanation!! :) – BitsOfNix Jun 6 '13 at 8:14

You don't need ls and grep for this; you can use a simple glob /a/[0-9]*:

echo /a/[0-9]*
ls /a/[0-9]*
grep foo /a/[0-9]*

If you're using this in a script, beware that parsing ls output is a bad idea.

share|improve this answer

I agree with l0b0 grep is a bad idea here but anyway, here is an explanation of the issue and a workaround. On Solaris 10 and older, /bin/sh is an antiquated shell that shouldn't be used for anything but running legacy scripts. You really should use ksh, bash or /usr/xpg4/bin/sh instead.

The root cause here is ^ used to be the original way to specify a pipe in the early Unix times. Solaris /bin/sh inherited this archaelogical feature.

The workaround is then quite simple, just escape the caret one of these ways:

ls /a |grep \^[0-9]


ls /a |grep "^[0-9]"


ls /a |grep '^[0-9]'
share|improve this answer
The first one would fail if there were a ^0 and ^1 file in the current directory though. You beat me by 7 seconds on the issue's Bourne source ;) – Stéphane Chazelas Jun 5 '13 at 13:38
@StephaneChazelas Why would the first one fail ? I just checked and it worked just fine. – jlliagre Jun 5 '13 at 16:12
Because the shell would expand it to ls /a | grep '^0' '^1'. That is, it would search for lines starting with 0 in ^1 and ignore the output of ls. – Stéphane Chazelas Jun 5 '13 at 16:49
@StephaneChazelas Got it, I overlook the "and". Thanks. – jlliagre Jun 5 '13 at 17:03

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