Shell Programming and Scripting

Glenn,

If I need to through out an error , how should I set it Iam doing like this but it's not throughing error.

#!/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 echo failure
fi

If you want to be verbose:

Code:

if (( $filecount == 10 )); then
  print "success"
  exit 0
else
  print "failure"
  exit 1
fi

Glenn,

Thank you very much it's throughing an error now. But if I don't need to pass the whole name of the file but just BARE01_DLY.

some thing like this: check20.sh BARE01_DLY

But it has to check whether 20 files are present for that day or not? I am using the code below. It's not working. What's wrong in this?

#!/bin/ksh
myfilepattern=$@
integer filecount=0
for file in $myfilepattern_$(date +%Y%m%d); do
filecount=$filecount+1
done

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

Please suggest

Glenn,

I am getting the following error when I run the debug mode.

myfilepattern=BARE01_DLY
+ typeset -i filecount=0
./script3.ksh[7]: syntax error at line 9 : `(' unexpected

Glenn,

The code I modifed like this;

#!/bin/ksh


dir=/biddf/ab6498/dev/ctl
export dir
set -x
myfilepattern=$@
integer filecount=0
for file in $myfilepattern_???_(date +%Y%m%d); do
filecount=$filecount+1
done

if (( $filecount == 10 )); then
print "success"
exit 0
else
print "failure"
exit 1
fi

Please suggest

Glenn,

I removed the braces in between date field but i am surprised how the value of filecount is 2 here. See the debug output once I remove the braces between date field in the script.

myfilepattern=BARE01_DLY
+ typeset -i filecount=0
+ filecount=0+1
+ filecount=1+1
+ let 2 == 10
+ print failure
failure
+ exit 1

Here is the script;

#!/bin/ksh


dir=/biddf/ab6498/dev/ctl
export dir
set -x
myfilepattern=$@
integer filecount=0
for file in $myfilepattern_???_date +%Y%m%d ; do
filecount=$filecount+1
done

if (( $filecount == 10 )); then
print "success"
exit 0
else
print "failure"
exit 1
fi

I have a script like this below. I am calling the script like this script1.ksh BARE01_DLY The script looks in the directory for named files BARE01_DLY_???_YYYYMMDD. The YYYYMMDD signify todays date when the script ran. It checks whether they are 10 files with that pattern for today and if they are then it shows succes otherwise failure. But in the directory I have only one file with name BARE01_DLY_MKT_20060720. But after I run the script in debug mode the output shows like it has 2 files with that name. Please see belwo for output.

myfilepattern=BARE01_DLY
+ typeset -i filecount=0
+ filecount=0+1
+ filecount=1+1
+ let 2 == 10
+ print failure
failure
+ exit 1
Here it shows 2==10 but I have only one file for that day. So where am i going wrong. Please suggest. Below is the script.

Code:

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

if (( $filecount == 10 )); then
print "success"
exit 0
else
print "failure"
exit 1
fi