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Need more better and efficient solutions for useing regular expressions to validate the user input in csh?

I need some comments to get a more better and efficient solution for validating the user input in a csh script.

Here are my solutions

#last two digits of year (00..99)
if ( `echo $Yr | grep '^[0-9][0-9]$'` == '' ) then
echo INVALID Yr;exit 1
endif

#day of year (001..366)
if ( `echo $Doy | grep '^[0-2][0-9]\{2\}$'` == '' && \
`echo $Doy | grep '^3[0-6]\{2\}$'` == '' || \
$Doy == '000' ) then
echo INVALID Doy;exit 1
endif

#4-character site ID (can contain only letters, numbers)
if ( `echo $Site | grep '^[a-zA-Z0-9]\{4\}$'` == '' ) then
echo INVALID SiteID;exit 1
endif

#1-character session ID (can contain only 0 or letters a-x)
if ( `echo $SessID | grep '^[0|a-xA-Z]$'` == '' ) then
echo INVALID SessID;exit 1
endif

Any comments will be appreciated

Thanks

Dragon_Krome

* For the first expression, you can also use:
^[0-9]{2}$ , it's shorter

* For the day of year:

(if you want to take into account the leap years, you'd need to do more calculations)

if you want 1 ... 366 format you can use this regular expression:

^(36[0-6]|3[0-5][0-9]|[1-2][0-9]{2}|[1-9][0-9]|[1-9])$

This would allow more accurate checking of the day.

* The expression for the 1 char ID is wrong:
^[0|a-xA-Z]$ ---> change to this: ^[0a-xA-X]$

Jerryleo

ASKER

Thanks for reply

>^(36[0-6]|3[0-5][0-9]|[1-2][0-9]{2}|[1-9][0-9]|[1-9])$

It doesn't work. for example,
[jerry@beks-m04 ~]$ echo 203 | grep '^(36[0-6]|3[0-5][0-9]|[1-2][0-9]{2}|[1-9][0-9]|[1-9])$'
[jerry@beks-m04 ~]$

ASKER CERTIFIED SOLUTION

Dragon_Krome

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