Shell Programming and Scripting

Glenn,

You made a very good point. But here there will be 20 files for BARE01_DLY_MKT_YYYYMMDD alone and 20 files for BARE02_DLY_MKT_20060803 alone and like that with some more names. So that is why I need to pass BARE01_DLY_MKT_YYYYMMDD as the parameter so that it checks whether that named file has 20 files or not and similarly for other file names. Please remember that MKT is the only that changes in the file. So please help.

It's really the same concept... just pass the entire pattern instead of just the date.

check20.sh:

Code:

#!/bin/ksh
myfilepattern=$1
integer filecount=0
for file in $myfilepattern; do
  filecount=$filecount+1
done

if (( $filecount == 20 )); then
  exit 0
else
  exit 1
fi

Syntax:

check20.sh BARE01_DLY_???_20060803
check20.sh BARE02_DLY_???_20060803
check20.sh BARE01_DLY_???_20060720
etc.

Glenn,

I tried this and the code is not returning error if the file count doesn't match the required count. Please suggest.

Woops! Use $@ instead of $1.

Code:

#!/bin/ksh
myfilepattern=$@
integer filecount=0
for file in $myfilepattern; do
  filecount=$filecount+1
done

if (( $filecount == 20 )); then
  exit 0
else
  exit 1
fi

Glenn,

I ran the code and i am using set -x to debug the code and surprised by the way code is taking. I have created one file named BARE01_DLY_MKT_20060803 in the directory /biddf/ab6498/dev/ctl direcotry and the script should show an error but please see below for the debug statments from the script;

script2.ksh BARE01_DLY_MKT_20060803
+ myfilepattern=BARE01_DLY_MKT_20060803
+ typeset -i filecount=0
+ filecount=0+1
+ let 1 == 10
+ exit 1

It is taking as 1=10 and exit ing from the code with out througing an error. Below is the script I modified.

#!/bin/ksh
dir=/biddf/ab6498/dev/ctl
export dir
set -x
myfilepattern=$@
integer filecount=0
for file in $myfilepattern; do
filecount=$filecount+1
done

if (( $filecount == 10 )); then
exit 0
else
exit 1
fi

Please suugest

Glenn,

I don't need to pass the whole name of the file but just BARE01_DLY would suffice. But it has to check whether 20 files are present for that day or not?


Please suggest.

The "let 1 == 10" line is not setting 1 equal to 10. It is comparing 1 (the number of files it counted) and 10 (the number of files needed for success). Because 1 != 10, the script exits status 1, which is an error. The unix convention is that "0" is success, and anything else is failure.