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Passing commandline argument to a function

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# 1  
Old 03-15-2010
Passing commandline argument to a function
Hi,

I have 2 ksh scripts.
Script1.ksh contains function definition.

script1.ksh
Code:
function f1() {

while getopts a:c: args
do
case $args in
a) ARG1=$OPTARG ;;
c) ARG2=$OPTARG ;;
\?) echo "Error no valid Arguments passed"
esac
done
echo $ARG1
echo $ARG2

script2.sh
Code:
. script1.ksh

sql_query="select * from emp"

f1 -a "ABCD1234" -c $sql_query

When I execute the script2.ksh, I am getting output as ABCD1234 and select
Where I should get the select * from emp.
Can any one please let me know where is the error?

Thanks,
Siba
# 2  
Old 03-15-2010
Just use the follows.
Code:
f1 -a "ABCD1234" -c "$sql_query"

Here the variable $sql_query is having the string containing the space.
So you should use double quotes like "$sql_query".

Then only it will preserve the white space.
Now I ran the script and got the output as follows

Code:
ABCD1234
select func.sh sc.sh from emp

Cheers!!!
# 3  
Old 03-15-2010
Turn off wildcard expansion before you call script1.ksh and use double quotes and you should be fine.

script2.ksh
Code:
#!/usr/bin/ksh

#
# disable filename substitution.
#
set -o noglob

. script1.ksh

sql_query="select * from emp"

f1 -a "ABCD1234" -c "$sql_query"

#
# enable filename substitution.
#
set +o noglob

sql_query="select * from emp"

f1 -a "ABCD1234" -c "$sql_query"

test:
Code:
$ ./script2.ksh
ABCD1234
select * from emp
ABCD1234
select script1.ksh script2.ksh from emp

Good luck.
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