script not working...select utility


 
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# 1  
Old 05-08-2008
script not working...select utility

#!/usr/bin/bash


name="$@"


myname=malay
#echo $myname

select firstname in $name;
do
if [$firstname!= $myname];then
echo $firstname
else
break
fi
done

invoking with:-
./script.sh one two three four five six seven eight nine malay
according to me the program control should come out when i will select the entry having my name[malay] but its not happening

can any one point out the error
# 2  
Old 05-08-2008
Hi,

Syntax error is there in the test condition. space must be there after the opening bracket and bfore the closing bracket, like this

if [ $firstname != $myname ];then

Thanks
Penchal
# 3  
Old 05-08-2008
put this line at the beginning of the script and debug it.

Code:
set -x

# 4  
Old 05-08-2008
ash-3.00$ ./script.sh one two three four malay
one two three four malay
malay
1) one
2) two
3) three
4) four
5) malay
#? 1
./script.sh: line 12: [: one!=: unary operator expected
bash-3.00$

this is the error coming
# 5  
Old 05-08-2008
what are the changes that you made to the script
# 6  
Old 05-08-2008
make the changes as penchal suggested it will work fine for you..
There should be space as he suggested.

Thanks
# 7  
Old 05-08-2008
okay i got hold of the error

[ $firstname != $myname ]

...i was missing the spaces in between


thanks everyone who cared to help
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