Shell Programming and Scripting

Anu thanks a lot , I got a clue to start with.
I have 17 file and each having differnt mandatory column can i generalized the
man_col="1,2,3,4,5,6,10,11" , I mean i want to pass the main_col value from a function and man_col value can be changed in run time like it can be 1,3,6,9,5 or it can be any thing ?

while executing this script in ksh i got
the error
awk: syntax error near line 1
awk: bailing out near line 1

Also thanks to sayonm , but synom what you guess is right but i could not get what you mention in the last line "save the labour of changing the file name and the no ("25")...."

@jambesh

what i meant was , everytime to check a particular co, u have to change the col name....so i said that to save that labour, a basic script can be written that will do that by its own...somwht the same thing that u mentioned in the last but one reply to anbu

cheers,
sayon

ps: do let us know if u want any other help regarding the automation

Try this jambesh..is that what you are looking for...
you can run this with first argument as filename and from second onwards mandatory column numbers like
scriptname filename.csv 1 2 3 5 7

Code:

# Scriptname filename col1 col2 col3 .....
if [ ! -f "$1" ];then
echo "file $1 does not exist " && exit 1
fi
filename=$1
shift 1
set -A com $*
lineno=1
while read line
do
   i=0
  while [ $i -lt $# ]
  do
  val=`echo $line| cut -d ',' -f${com[$i]}`
  [[ -z "$val" ]] && echo "resend the file line $lineno column ${com[$i]} is null" && exit 1
   i=`expr $i + 1`
  done
lineno=`expr $lineno + 1`
done<$filename

Hi dhruv what is set -A com $* ? and when u write -> while [ $i -lt $# ] what is $# here, is it no of argument ? for what no of argumetn u r making comparison ? please tell me ! bit confused

what this program doing is taking all the column numbers in a array
set -A com $* will set com array with values which you passed in comman line
like sc.sh file.csv 1 4 7 8 then statement
set -A com $* will set com array with com[0]=1,com[1]=4,com[2]=7.......

while [ $i -lt $# ] here $# is number of arguments minus 1 and it will be only no of columns.because after getting filename we used shift 1
filename=$1
shift 1
set -A com $*

need any other clarification?

Hi Dhruv

You are Great boss . Working fine this one .
Can you please give me a clue on the other one tht i post also.