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validate date pattern using Regular Expression

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# 1  
Old 04-19-2010
validate date pattern using Regular Expression
Hi,
i am java guy and new to unix.
I want to validate date pattern using Regex expression
here is the sample program i have written.

Code:
#!/bin/sh
checkDate="2010-04-09"
regex="\\d{4}-\\d{2}-\\d{2}\$"
echo $regex
if [[ $checkDate =~ $regex ]]
then
    echo "OK"
else
    echo "not OK"
fi

But the ouput is "not Ok"
Please help me.
Last edited by pludi; 04-19-2010 at 04:58 AM.. Reason: code tags, please...
# 2  
Old 04-19-2010
You could try this with standard globbing:
Code:
#!/bin/bash or #!/bin/ksh93
checkDate="2010-04-09"
regex="[1-9][0-9][0-9][0-9]-[0-1][0-9]-[0-3][0-9]"
echo $regex
if [[ $checkDate =~ $regex]]
then
    echo "OK"
else
    echo "not OK"
fi

Last edited by Scrutinizer; 04-19-2010 at 04:25 PM.. Reason: Changed glob to regex. Somehow thought globs were used for regex, but that is incorrect. Tnx Snemetz
# 3  
Old 04-19-2010
Fixed regex
Change your regex to

Code:
regex='[0-9]{4}-[0-9]{2}-[0-9]{2}$'

\d is part of the extended regex set, which bash does not support by default.
There may be a way to get it to use the extended set, but I've haven't checked.
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