array element compare in while condition

[drewbyh]

The following is part of a script I'm writing to show releases of software available in a directory and then ask the user to select one. While $answer is not in the avail[] array ask the question again. I need to know how to compare each element in the array to the answer variable in the while loop condition. Any help is greatly appreciated. Thanks

avail=(`ls -C`)
echo "Releases available: ${avail[*]}"

echo -n "What SCSRelease whould you like to run? (x.x.x) "; read answer

# This part does not work

while [ "$answer" != "${avail[*]}" ];
do
echo -n "Release is already running or not valid. Try again: (x.x.x) "
done

 

[ericbrunson]

You'll probably have to write a function that iterates over the array values, comparing the answer to each one and return success when it succeeds.

 

[vgersh99]

or.... use the 'select' clause: 'man ksh' and search for 'select'.

Not exactly using the arrays, but building a 'menu-driven' interface - which what I think you're trying to do.

vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[PHV]

In your example, why not simply test the existence of the file ?
while [ ! -f "$answer" ]

Hope This Helps, PH.
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