Need to monitor a dir for files
 

Need to monitor a dir for files
if count > 0, sleep for 3min and check again for count and if count >0 then write to logfile.

if count = 0 then write to logfile(no files available)

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OK. Not sure what your question is...

What have you tried?
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Hi
I tried the below

#!/bin/bash
dir="/tmp/test1/test"
LOG_FILE="/tmp/test1/files_mon.log"
>$LOG_FILE
count=`find $dir -type f -ls | wc -l`
if [ $count -ge 1 ]; then
sleep 10s
count=`find $dir -type f -ls | wc -l`
if [ $count -ge 1 ];
then
echo -e "Error : Files are not processed under dir $dir " >>$LOG_FILE
else
echo "INFO : All files are processed under dir $dir">>$LOG_FILE
fi
fi

That script looks fine... except you don't deal with the == 0 case. What is the problem?

Please use [code] tags as above when posting code or output

Does that script not work? It looks like it should, except it doesn't match the specs in your OP. The sleep should be 180s or 3m (shorter for testing, perhaps?)

What is supposed to be happening in the directory that would change the count? If that's not happening, you'll never get the else. There are but two tests in 10 secs, then the script is done.

wouldn't it be better to use inotifywait/inotifywatch for this?

Quite possibly. But if it really does take minutes for the files to be processed, the script as shown is pretty simple with no real parsing necessary.