Parameters in loop

Shell Programming and Scripting
1

Hi,

I am trying to write a script which will read inputs form user and process those files, I have issue reading the input parameters in a loop. Following is the script...

I run the script as ./Script.sh 3 table1 table 2 table3

NumberOfTables=$1
let TableCount=1

while [ ${NumberOfTables} -gt 0 ]
do

 TableName='$'$TableCount
 
 db2 "runstats on table ${TableName} and indexes all"

 let  TableCount=TableCount+1
 let  NumberOfTables=NumberOfTables-1

done
exit 0

here I am not able to capture table1 table2 and table3 in the loop it prints TableName as $1 $2 and $3 but not the names that are given as input.

can some one help me on this....

2

Having trouble following your logic, but perhaps the following thoughts will help this move along:
(1) The variable "$#" will be the number of parameters supplied. Thus, perhaps no need for the first 3 after your command because you could skip this and then do NumberOfTables="$#"
(2) The shift function allows a script to keep processing references to $1. Thus, you would do your first set of commands and then do a shift. Shift moves what is in $2 to $1, $3 to $2, etc... ; what was in $1 is now gone.

3

I do not think you need the TableCount variable, and perhaps a few other things inside your original code. Hopefully, this will put you on your way...

> cat script.sh
#! /bin/bash

NumberOfTables="$#"
let TableCount=1

while [ ${NumberOfTables} -gt 0 ]
do

 TableName='$'$TableCount
 
# db2 "runstats on table ${TableName} and indexes all"
# echo ${TableName}
 echo $1
 shift 

 let  TableCount=TableCount+1
 let  NumberOfTables=NumberOfTables-1

done
exit 0
4

Thanks It helped a lot...