Counting files in a directory that match a pattern - Page 5 - Shell Programming and Scripting

aigles · #29 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 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.

dsravan · #31 07-24-2006

Jean,

That works perfect!! I just got a question don't the condition need to be reverse of what we have in the for loop condition. I marked it in red. Kindly suggest.

#!/bin/ksh -x
dev='/biddf/ab6498/dev/ctl'
cd $dev
echo $dev

Quote:

#set -x
myfilepattern=$@
fullpattern=${myfilepattern}_???_$(date +%Y%m%d)
integer filecount=0
for file in $fullpattern ; do
[ "$file" != "$fullpattern" ] && filecount=$filecount+1
done

if (( $filecount == 2 )); then
print "success"
exit 0
else
print "failure"
exit 1
fi

dsravan · #32 07-24-2006

Jean,

I am sorry. I got why we should have negate condition. Thanks for the help.

Once again thanks for your great help.

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