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Full Discussion: Passing argumnets from shell script to sql
Top Forums Shell Programming and Scripting Passing argumnets from shell script to sql Post 302142045 by bakunin on Wednesday 24th of October 2007 04:38:14 AM
Old 10-24-2007
Your commandline arguments show up in the script as "$1", "$2", etc. If you have more than 9 arguments ($1..$9) you must use "shift" to get all of them.

There is also a special variable "$#", which holds the number of passed arguments and "$*" which holds all passed arguments.

shift will make $9 to $8, ..., $2 to $1 and lose the previous content of $1. $9 which is freed this way will hold the 10th passed argument if there is one.

Here is a little demonstration script to show you the mechanism:

Code:
call the following script with different numbers of arguments:

script.sh a b c d
script.sh "a b" c d e f g h
etc.

#! /bin/ksh

typeset    chArg1=""     # a normal variable

print - "The number of commandline arguments was: $#"
print - "all arguments: $*"

chArg1="$1"             # store the first argument in chArg1

print - "you can transfer the arguments to normal variables:"
print - "the first argument directly: $1"
print - "and here the content of chArg1: $chArg1"

print - "we now see the shift-command in action"

while [ $# -gt 0 ] ; do
     print - "Next argument..: $1"
     print - "all arguments...: $*"
     print - "Nr of arguments.: $#"
     print - "------------"
     shift
done

exit 0

With these mechanisms you should be able to not only put your arguments into your SQL-Statements, but also validate your script to recognize too few/too many or syntaktically wrong arguments.

bakunin
 

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rt-shredder(8)						     Request Tracker Reference						    rt-shredder(8)

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