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# 1  
Old 05-06-2012
bash pattern
hi!

When I have a variable that contains the day let's say Sun06
and i want to check if my variable starts with any letter what pattern should I use?

I tried that one,but it returns syntax error..
Code:
if [ "$day" = ([A-Za-z]*) ]; then

# 2  
Old 05-06-2012
I would use case instead of if:
Code:
case "$day" in
([A-Za-z]*) echo ok ;;
(*) echo ko ;;
esac

This User Gave Thanks to jlliagre For This Post:
vlm
# 3  
Old 05-06-2012
I would also have a preference for case since it is more portable. Since you are using bash you could also use:
Code:
if [[ "$day" == [A-Za-z]* ]]; then

--
@jlliagre: in light of our other interesting thread, the double quotes can be left out around the variable in the case statement ( case $day in )
This User Gave Thanks to Scrutinizer For This Post:
vlm
# 4  
Old 05-07-2012
Quote:
Originally Posted by Scrutinizer
I would also have a preference for case since it is more portable. Since you are using bash you could also use:
Code:
if [[ "$day" == [A-Za-z]* ]]; then

--
@jlliagre: in light of our other interesting thread, the double quotes can be left out around the variable in the case statement ( case $day in )
it can also be left out the left hand side in bash's [[ Smilie

Code:
[[ $day = [[:alpha:]]* ]]

These 2 Users Gave Thanks to neutronscott For This Post:
Scrutinizer vlm
# 5  
Old 05-07-2012
Ow, yes of course Smilie Thanks neutronscott..
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