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> That problem is solved. Good. > > My next question is related to ISL scripting. > > 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" > I did not document this in the ISL docs (I am now) but along with the file property there is also a filetail and a fileprefix property. e.g.: ${file} is the full filename, e.g.,. /test/data/example.nc ${filetail} is the file name: example.nc ${fileprefix} is the full path with the suffix stripped off, e.g.: /test/data/example I see I should have a property which is the fileprefix without the dir path. -Jeff Ticket Details =================== Ticket ID: IBB-842776 Department: Support IDV Priority: Normal Status: Closed