Hi,
I have a list file which has script names in them. Some scripts take lists as parameters which inturn have script names. This is basically for sequencing the job run.

Eg: List1:

Code:

test1.ksh
test2.ksh test2.lst
test3.ksh test3.lst
test4.ksh

test2.lst:

Code:

test21.ksh
test23.ksh
test23.ksh

test3.lst:

Code:

test31.ksh
test33.ksh
test33.ksh

I need to write code to create one master list from this which will look like:

Code:

test1.ksh
test21.ksh
test23.ksh
test23.ksh
test31.ksh
test33.ksh
test33.ksh
test4.ksh

I am trying to grep every line in List1, if it contains .lst get the list file name and echo the contents to create the master list.
It gives me an error as the line contains a script name and the script is not in the directory.
Is there a wat I can grep a line in a file which contains a script name?

Code:

#!/bin/ksh
while read line
do
   echo $line
   sublst_name1=`echo $line | cut -d " " -f2`
   echo $sublst_name1
   lst_cnt=$(grep -i .lst $sublst_name1 | wc -l)
   echo $lst_cnt
done <"$lst_file"

output:

Code:

test1.ksh
test1.ksh
grep: can't open test1.ksh
0
test2.ksh test2.lst 
test2.lst
2
test3.ksh test3.lst 
test3.lst
2
test4.ksh
test4.ksh
grep: can't open test4.ksh
0

Thanks in advance for the help.

Moderator's Comments:

Please use code tags!