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Argument check

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# 1  
Old 08-17-2014
Argument check
All,

I'm writing an argument checking block and running into an error. I want to confirm that $1 is one of two values. Here is what I have:

Code:
if [[ "$1" != ("-e" || "-d") ]]; then

        echo -e "\nPlease check your first augument. You used \"$1\"
which is not recognized.  Please see usage:"
        usage
        exit 1

else

        echo -e "\narg 1 is good to go.."

fi

Can anyone point me in the right direction? Any help is greatly appreciated.

Thanks,

Herb
# 2  
Old 08-17-2014
The correct syntax depends heavily on the shell that you are using. Please post. What would you conclude from the error message that you certainly got but did not post either?
# 3  
Old 08-17-2014
Close, try changing your if condition to the following:

Code:
if [[ "$1" != @("-e"|"-d") ]]; then

These 2 Users Gave Thanks to pilnet101 For This Post:
hburnswell RudiC
# 4  
Old 08-17-2014
RudiC - My apologies, I'm writing in bash.

pilnet101 - I made your suggested modification as:

Code:
if [[ "$1" != @("-e"|"-d") ]]; then

        echo -e "\nPlease check your first augument. You used \"$1\"
which is not recognized.  Please see usage:"
        usage
        exit 1

else

        echo -e "\narg 1 is good to go.."

fi

I receive the following error:

Code:
syntax error in conditional expression: unexpected token `('
syntax error near `@("'
`if [[ "$1" != @("-e"|"-d") ]]; then'

Did I write it incorrectly?

TIA,

Herb
Last edited by Scrutinizer; 08-17-2014 at 05:18 PM.. Reason: additional code tags
# 5  
Old 08-17-2014
Surprisingly enough, pilnet101's proposal worked in my GNU bash, version 4.3.11(1). I'm glad I learned something today (although a bit disappointed as I didn't find that feature in bash's man page)!

As your bash doesn't like that construct, try e.g. (amongst other options):
Code:
if [[ "$1" != "-e" && "$1" != "-d" ]]; then

Last edited by RudiC; 08-17-2014 at 03:05 PM.. Reason: typo
This User Gave Thanks to RudiC For This Post:
hburnswell
# 6  
Old 08-17-2014
The extended pattern became default behavior inside double square brackets in bash version 4.1, so I imagine OP's bash version is an older one.
# 7  
Old 08-17-2014
Thanks RudiC. Yes my bash version is quite older: GNU bash, version 3.2.25(1).

I was able to get the desired behavior with your suggested modification.

I appreciate your help.

Herb
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