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problem with getopts

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Top Forums Shell Programming and Scripting problem with getopts
# 1  
Old 07-23-2012
problem with getopts
Hi,

I have written a script to take command line arguments using geopts.This is the code.

Code:
#!/bin/sh
# The usage of this script.
usage="Usage is $0"
usage="$usage [-a <database>]"
usage="$usage [-b <parameter>]"
usage="$usage [-c <table_name>]"
# Use the getopt utility to set up the command line flags.
set -- `/usr/bin/getopt a:b:c: $*`
echo "$?"
if [ $? -ne 0 ];then
  echo "Invalid script invocation"
  echo $usage
  exit
fi
# Process individual command line arguments
while [ $1 != -- ]; do
  case $1 in
    -a)  database=$2
         shift
         ;;
    -b)  parameter=$2
         shift
         ;;
    -c)  tablename=$2
         shift
         ;;
  esac
  shift
done

echo "$database"
echo "$parameter"
echo "$tablename"


When i am excuting this script without passing any parameter then it is not showing invalid invocation as it should. Also when i am not passing any parameter then also set command is returning value as 0 means success, but it should return 1 as there is no parameter passed to set means failure. PLease see the execution results of the script.

Code:
ksh -x test.sh
+ usage='Usage is test.sh'
+ usage='Usage is test.sh [-a <database>]'
+ usage='Usage is test.sh [-a <database>] [-b <parameter>]'
+ usage='Usage is test.sh [-a <database>] [-b <parameter>] [-c <table_name>]'
+ /usr/bin/getopt a:b:c:
+ set -- --
+ echo 0
0
+ [ 0 != 0 ]
+ [ -- != -- ]
+ echo ''
+ echo ''
+ echo ''

Please suggest. Is there something wrong in the code.Smilie

---------- Post updated at 01:25 PM ---------- Previous update was at 01:24 PM ----------

This is the code that i am executing.

Code:
#!/bin/sh
# The usage of this script.
usage="Usage is $0"
usage="$usage [-a <database>]"
usage="$usage [-b <parameter>]"
usage="$usage [-c <table_name>]"
# Use the getopt utility to set up the command line flags.
set -- `/usr/bin/getopt a:b:c: $*`
echo "$?"
if [ $? != 0 ];then
  echo "Invalid script invocation"
  echo $usage
  exit
fi
# Process individual command line arguments
while [ $1 != -- ]; do
  case $1 in
    -a)  database=$2
         shift
         ;;
    -b)  parameter=$2
         shift
         ;;
    -c)  tablename=$2
         shift
         ;;
  esac

shift
done
Last edited by zaxxon; 07-23-2012 at 05:36 AM.. Reason: guess what
# 2  
Old 07-23-2012
What are you trying to do here?
Code:
 `/usr/bin/getopt a: b:c: $*`

# 3  
Old 07-23-2012
Hi,

My script requires 3 command line arguments to execute it.
If someone does pass these 3 arguments then it should return that invalid invocation and the usage.But in my script even if i do not pass any argument it is not going in that if part. Also when i am not passing any argument then also set is returning the value as 0 .i.e. successfull execution instead of 1 .i.e failure. Since it is returning 0 it is not going into that if part where i have mentioned if [$? -ne 0];then echo "something".

Please suggest.Smilie
# 4  
Old 07-23-2012
Maybe you could use $# for the number of arguments passed!!

That should work fine! Sorry I am lil' confused with this getopt idea!
# 5  
Old 07-23-2012
Code:
$# also didnot worked.
 
ksh -x test.sh
+ usage='Usage is test.sh'
+ usage='Usage is test.sh [-a <database>]'
+ usage='Usage is test.sh [-a <database>] [-b <parameter>]'
+ usage='Usage is test.sh [-a <database>] [-b <parameter>] [-c <table_name>]'
+ /usr/bin/getopt a:b:c: 0
+ set -- -- 0
+ echo 0
0
+ [ 0 -ne 0 ]
+ [ -- != -- ]
+ echo ''
+ echo ''
+ echo ''

still it is not going into that if block.
With $# it is showing 0 arguments passed, but still not going into the if block.Smilie
Last edited by Scott; 07-23-2012 at 09:42 AM.. Reason: Deja vu
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