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The correct syntax would be something something like this, using arithmetic evaluation:
if (( dm > 0 || cl == 100 || ad == 0 || me > 8 || cp > 0 )); then

or if you want to use the test utility,...

Have you investigated the logrotate command and configuration? That might save you having to craft your own. You can force a rotation based on several criteria as it suits your needs.



Robin

try adding to cron:
00 00 * * * test $(( ($(date +\%s)/24/60/60) \% 45 )) = 0 && /path_to_script/clean_up_script

ever considered using at command?
The idea in your job execution, to do self execution by calling itself in 45 days using at...
I dp that sometimes...

See updated post #2.

With bash, try
printf "%0${#LOGNUM}d\n" "$(( 10#$LOGNUM - 1 ))"
0000094

When you post questions like this in the future, please tell us what operating system and shell you're using!

If you're using ksh, the easy way to do it is:
LOGNUM=0000095
OLDLOG=$(printf "%07d"...

I doubt that it is working for all numbers below 99. And, you haven't told us what shell you're using to get these results. The above output is exactly what I would expect if you're using bash as...

LOGNUM=0000095
printf "%0${#LOGNUM}d\n" $(echo "$LOGNUM-1" | bc)

Does it have to be bc? Try
awk -vLNUM=$LOGNUM 'BEGIN {printf "%07d\n", LNUM-1}'
0000094

find . mtime +2 -type f \( -name '*.LOG' -a ! -name S0000000.LOG \) -exec rm {};

Assuming I got what you want try this. Just do not run the exec part until you are sure it does what you want....

It all depends what you are going to do with it. As an example this does:

perl -nle '/CURDEPTH\((\d+)\)/ and print $1' prince1987.file

Edit: Added a + to accommodate that a number could be more...