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Bash - check the first letter of a var

Posted on 2007-04-06
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Last Modified: 2010-05-18
while [ -z $cnickname ] || echo ${cnickname:0:1} | grep [:digit:];
do
if echo ${cnickname:0:1} | grep [:digit:]; then
something

When i execute this it says: Syntax error: Bad substitution, any ideea whats wrong ?
The script should take the virst letter of $cnickname and see if its a digit, and if it is do something
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Question by:rares_dumitrescu
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4 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 18863049
Are you sure you are running it in bash and not in sh?
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Author Comment

by:rares_dumitrescu
ID: 18863093
you're right i was using /bin/sh, i changed it to #!/usr/local/bin/bash .. now it works but the if, or while does not turn on at echo ${cnickname:0:1} | grep [:digit:]; when i put something like '3blah'
0
 

Author Comment

by:rares_dumitrescu
ID: 18863113
worked
i know i might sound very stupid, but, why didn't it worked with /bin/sh and it works with /bin/bash ?
whats the difference ? (i am making some scripts on a server, like getegg getpsy etc (for irc) using dialog menu)
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LVL 84

Accepted Solution

by:
ozo earned 500 total points
ID: 18863190
they are different shells, see
man sh
and
man bash
you might try grep '[0-9]'
or
grep '[[:digit:]]'
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