Counting files in a directory that match a pattern - Page 2 - Shell Programming and Scripting

Glenn Arndt · #16 (**permalink**) 07-20-2006

If you want to be verbose:

Code:

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

dsravan · #17 (**permalink**) 07-20-2006

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

dsravan · #18 (**permalink**) 07-20-2006

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

dsravan · #19 (**permalink**) 07-20-2006

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

dsravan · #20 (**permalink**) 07-20-2006

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

dsravan · #21 (**permalink**) 07-20-2006

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

reborg · #22 (**permalink**) 07-20-2006

threads merged, please do not start multiple threads for the same topic.

dsravan · #23 (**permalink**) 07-20-2006

Reborg,

I am sorry for that!!! I got no answer so thought that new thread may incerase people viewing.

Anyway can you help me with this please?

reborg · #24 (**permalink**) 07-20-2006

Actually that exactly the reason that it is againts the rules of these forums.

However try this:

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

dsravan · #25 (**permalink**) 07-20-2006

Reborg,

I bow to you completely. You are great. It's working great man.

I will learn from this thread a lot. I will be active member from now on.

Once again thank you very much.

dsravan · #26 (**permalink**) 07-24-2006

Hi,

I have a problem with the script below. Basically it checks whether there is any file present with the name CARE01_DLY present for today's date in the directory /biddf/ab6498/dev/ctl and if the count =1 it says success otherwise error. But it doesn't seem to take the todays date file but checking if the file is present and saying success if the file is present with out checking for todays date. But i want the script to check for todays file only and if found it should through success otherwise error. The name of the file is sent as an argument to the script.

Quote:

#!/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

here is the output: The script is called like this script3.ksh BARE02_DLY

Quote:

+ dir=/biddf/ab6498/dev/ctl
+ export dir
+ set -x
+ myfilepattern=BARE02_DLY
+ typeset -i filecount=0
+ date +%Y%m%d
+ filecount=0+1
+ let 1 == 1
+ print success
success
+ exit 0

There is no file with this name for today but it's saying success.

Please help.

aigles · #27 (**permalink**) 07-24-2006

I found two problem in your script

$myfilepattern_???_$(date +%Y%m%d) is interpreted by the shell as ${myfilepattern_}???_$(date +%Y%m%d).
You must code like this : ${myfilepattern}_???_$(date +%Y%m%d)

The for statement returns the pattern itself if no file are found.
You must test for that pattern.

Code:

#!/bin/ksh
dir=/biddf/ab6498/dev/ctl
export dir
set -x 
myfilepattern=$@
fullpattern=${my_filepattern}_???_$(date +%Y%m%d)
integer filecount=0
for file in $fullpattern ; do
   [ "$file" != "$fullpattern"] && filecount=$filecount+1
done

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

Jean-Pierre.

dsravan · #28 (**permalink**) 07-24-2006

Jean,

Thanks for the changes. But the script is still producing wrong result. There are 2 files with todays date in the directory and still it says failure. I think the main problem is pattern matching is is complaring like this:

CARE01_DLY_AUS_20060724 != CARE01_DLY_???_20060724 and saying no match. So please suggest.

Please see output:

Quote:

+ dev=/biddf/ab6498/dev/ctl
+ cd /biddf/ab6498/dev/ctl
+ echo /biddf/ab6498/dev/ctl
/biddf/ab6498/dev/ctl
+ myfilepattern=CARE01_DLY
+ + date +%Y%m%d
fullpattern=CARE01_DLY_???_20060724
+ typeset -i filecount=0
+ [ CARE01_DLY_AUS_20060724 != CARE01_DLY_???_20060724]
./script4.ksh[11]: test: ] missing
+ [ CARE01_DLY_MKT_20060724 != CARE01_DLY_???_20060724]
./script4.ksh[11]: test: ] missing
+ let 0 == 1
+ print failure
failure
+ exit 1

aigles · #29 (**permalink**) 07-24-2006

Quote:

+ dev=/biddf/ab6498/dev/ctl
+ cd /biddf/ab6498/dev/ctl
+ echo /biddf/ab6498/dev/ctl
/biddf/ab6498/dev/ctl
+ myfilepattern=CARE01_DLY
+ + date +%Y%m%d
fullpattern=CARE01_DLY_???_20060724
+ typeset -i filecount=0
+ [ CARE01_DLY_AUS_20060724 != CARE01_DLY_???_20060724]
./script4.ksh[11]: test: ] missing
+ [ CARE01_DLY_MKT_20060724 != CARE01_DLY_???_20060724]
./script4.ksh[11]: test: ] missing
+ let 0 == 1
+ print failure
failure
+ exit 1

Sorry, type error
Add a space befor the ] in the test statement whithin the for loop

Code:

#!/bin/ksh
dir=/biddf/ab6498/dev/ctl
export dir
set -x 
myfilepattern=$@
fullpattern=${my_filepattern}_???_$(date +%Y%m%d)
integer filecount=0
for file in $fullpattern ; do
   [ "$file" != "$fullpattern" ] && filecount=$filecount+1
done

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

Glenn Arndt · #30 (**permalink**) 07-24-2006

You need a space between the " and the ] in:

Code:

[ "$file" != "$fullpattern"]

Perhaps it would be easier for everyone if you familiarized yourself with the syntax of the shell you are using, rather than going back and forth dozens of times about tiny syntax problems. I'm not trying to be rude, just thinking that spooning tiny little pieces to you is less helpful than your taking the time to understand the general concepts of the environment in which you are programming. I've found that many of the experts here tend to give general solutions as a starting point to help you understand your specific problems, which is much more productive than debugging the same script multiple times.

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