Possible can be done using python scripting?
-jun
-----Original Message-----
From: Stuart Wier [mailto:wier@xxxxxxxxxx]
Sent: Tuesday, April 06, 2010 10:20 AM
To: Jun Chen; idvdevelopers@xxxxxxxxxxxxxxxx
Subject: Re: [idvdevelopers] ISL scripting
Jun Chen wrote:
> Following that example isl file multiplefiles.isl:
>
> <isl offscreen="false">
> <fileset pattern=".*\.nc">
> <echo message="File: ${file}"/>
> <bundle file="${islpath}/test.xidv">
> <!-- This sets the data file in the bundle to the ${file} -->
> <setfiles datasource=".*" file="${file}"/>
> </bundle>
> <pause seconds="20"/>
> </fileset>
> </isl>
>
> I want to dynamically generate each image file based on each file
> ${file}, but that file contains path and link from a location I only
> have read permission, my question is:
>
> How can I extract the (only) file name (without file extension) from
> ${file}?
> Say ${file}="/myserver/mydirectory/myfile1.nc"
>
> Is there any way I can extract just myfile1 from ${file}?
>
> So I can add:
>
> <image file="${islpath}/${myfilename}.png" />
>
> Above, ${myfilename} should change corresponding to ${file}
>
> Please help.
>
> I am new to IDV, and found it's very useful, but poor documentation.
>
>
Use shell scripting to create your file names and then use the shell to
write your isl file dynamically.
ISL is not a replacement for the shell.
