Shell Programming and Scripting

I need to zip all the files within a directory on a daily basis and store them as a zip file within that directory with the date in its naming convention

I have file extentions .log .lst and .out's within that directory

How do i do this task

zip -mqj mydir/allzip$YYMMDDDD.zip mydir/*.*

my question is I cannot give a *.* in the end. because tomorrow i will have todays zip file and i dont want to include this into my today's zip file how do i accomplish this.

Any thoughts are appreciated

Regards
Ram

Ram,
See if this would work for you:

Code:

zip -mqj mydir/allzip$YYMMDDDD.zip `ls -1 mydir/*.log mydir*.out mydir/*.lst`

Try :

Code:

zip -mqj mydir/allzip$YYMMDDDD.zip mydir/*.log mydir/*.lst mydir/*.out

Jean-Pierre.

Code:

!#/bin/ksh
zip -mqj mydir/allzip$YYYYMMDD.zip mydir/*.@(.lst|.out|.log)

will select files with the three endings: .out .lst .log
Try it with ls on your system first.

Jim,
Great solution!
Since it already has a "." before the "@", it is not needed before the extensions, in other words, it would be like:

Code:

zip -mqj mydir/allzip$YYYYMMDD.zip mydir/*.@(lst|out|log)

Yes It Works a big thanks to everyone.

By the way how do i zip for files with yestedays time stamp only. I do not want to zip todays files .As i do not have a timestamp in the naming coventions of this log,lst and out files.....
How do i only zip files with yesterdays time stamp on them.

Regards
Ram