Error in shell script

hasmeet.singh337 · 03-07-2011, 01:42 AM

hi everybody,
I have created a shell script to check the input number as positive or negative integer and if no argument is found then it should display a message that you must input a argument..

Script is given as:

vi pntest
#!/bin/sh
#
#
#Script to check wether integer is positive or negative
#
#
if [ $# -eq 0]
then
echo "$0 : You must give or supply one integer"
exit 1
fi
if test $1 -gt 0
then
echo "$1 is a positive number"
else
echo "$1 is a negative number"
fi

commands executed on it
chmod 755 pntest
./pntest

output is:

./pntest: line 7: [: missing `]'
./pntest: line 12: test: -gt: unary operator expected
is a negative number

Ashkan_s · 03-07-2011, 01:49 AM

you need a spcae between 0 and ]

Code:

if [ $# -eq 0]

Code:

if [ $# -eq 0 ]

Telengard · 03-07-2011, 02:39 AM

I tried your program.

Code:

~/foo$ sh pntest
pntest : You must give or supply one integer
~/foo$ sh pntest 5
5 is a positive number
~/foo$ sh pntest -5
-5 is a negative number
~/foo$ sh pntest 0
0 is a negative number

You might want to reconsider whether or not 0 is truly a negative number. (Or is 0 technically not an integer?)

One solution might be to add a third else clause to check for 0 and respond appropriately.