Passing argumnets from shell script to sql

Shell Programming and Scripting
1

hi I all ,

I have sql statment in my shell script , I pass two argument to the script I need to pass the this two arguments to the sql statment

example :

runsql.sh "1" "2"

sql :

updat tables_x set y=0
where
A=:x should subsituted by "1"
and
B=:y shuold subsituted bt "2"

how I can do that

thanks

2

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:

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