KornShell - Test with variable that may be not set

Question

I have the following code in KornShell (ksh):

FAILURE=1 SUCCESS=0 isNumeric(){ if [ -n "$1" ]; then case $1 in *[!0-9]* | "") return $FAILURE; * ) return $SUCCESS; esac; else return $FAILURE; fi; } #... FILE_EXT=${FILE#*.} if [ isNumeric ${FILE_EXT} ]; then echo "Numbered file." fi #...

In some cases the file name not have an extension, and this causes the FILE_EXT variable to be empty, which causes the following error: ./script[37]: test: 0403-004 Specify a parameter with this command.

How should I be calling this function so that I do not get this error?

I would expect the whole filename to be in the FILE_EXT rather than nothing, unless the file name ends in '.' — John Weldon
– John Weldon, Commented Dec 11, 2009 at 22:05

Dennis Williamson · Accepted Answer · 2009-12-12 06:37:13Z

You should leave off the square brackets when you are testing the exit code of a function, otherwise, you'll always get "true". Also, you should quote your variable. You could put an additional test for an empty extension as shown:

FILE_EXT=${FILE#*.} if isNumeric "${FILE_EXT}" && [ "${FILE_EXT}" != "${FILE}" -a "${FILE_EXT}" != "" ] then echo "Numbered file." fi

Edit: added test to handle filenames that end in "."

John Weldon · Accepted Answer · 2009-12-11 22:08:00Z

I would do this:

if [ isNumeric ${FILE_EXT:="no"} ]; then echo "Numbered file." fi

If all you want to do is determine that the file has a numeric extension

The ${FILE_EXT:="no"} will expand to either the value of FILE_EXT, or 'no' if FILE_EXT is null or empty.

False positive for filenames that consist only of digits since $FILE_EXT will contain the full filename if there is no dot at all.

ghostdog74 · Accepted Answer · 2009-12-12 00:03:38Z

you should use ${FILE##*.} with double "#" instead. also what do you mean the variable $FILE_EXT will be empty? if your file don't have extension, then when you do ${FILE#*.} you will get the just file name in FILE_EXT. how is it empty ?

Henk Langeveld · Accepted Answer · 2011-03-20 02:26:03Z

Assuming ksh93, it should be possible to use it's own arithmetic. But we need to be careful: Just ((n)) will fail if n==0, so we test for ((n || !n)) which should always be true for any proper number.

To prevent ksh from exiting, we run the expression in a subshell ( ), adding spaces to prevent conflicts with the arithmetic expression (( ... )).

Finally, we close stderr with '2>& -' to prevent any error messages from non numeric arguments, although you might want to keep them in.

function isNumeric { ( typeset n=${1:?} ((n||!n)) ) 2>& - }

David R Tribble · Accepted Answer · 2009-12-11 22:05:24Z

[ -z "$1" ] will test for an empty $1, as will [ "" = "$1" ].

Or you could simply prepend a "0" to $1 (i.e, "0$1") to force it to be non-empty before checking if it is numeric (if you want empty extensions to be treated as numeric).

user224243 · Accepted Answer · 2009-12-11 23:17:10Z

I had some probs to run your script (Maybe that's because im using pdksh). So I have adjusted it slightly. Try this:

#!/usr/bin/ksh FILE=$1 FAILURE=1 SUCCESS=0 isNumeric () { if [ -n "$1" ] then case $1 in *[!0-9]* | "") echo "$1 not a number" return $FAILURE ;; * ) echo "$1 is a number" return $SUCCESS ;; esac else echo "parameter is empty" return $FAILURE fi } #... FILE_EXT=${FILE#*.} echo $FILE_EXT isNumeric "${FILE_EXT}" if [ "$?" = "0" ] then echo "Numbered file." fi

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