Re: Getting the file name in a package
- To: mathgroup at smc.vnet.net
- Subject: [mg52919] Re: Getting the file name in a package
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Fri, 17 Dec 2004 05:18:26 -0500 (EST)
- Organization: Uni Leipzig
- References: <cpri7j$qsd$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi, what may ContextToFileName[] do ? BTW a package can be in a notebook and does not need a file name. Regards Jens "Geronimo" <kalymereauKILLSP at Myahoo.fr> schrieb im Newsbeitrag news:cpri7j$qsd$1 at smc.vnet.net... > Hi > > I have a package (.m) that produces some output in an external file. As > the > computation is quite long, I would like to avoid it if the output file is > more recent than the last modification of the package (since the output > only depends on what is in the package). > > So I need to extract the file name of the package within itself (and then > test the date with FileDate). As the context name is related to the file > name, I use: > > > BeginPackage["TestPack`"] > > (blabla) > > Begin["`Private`"] > > currentFileName = StringDrop[Context[], -9] <> ".m" > > (blabla) > > End[] > > EndPackage > > Since within the Private context the Context[] command gives > TestPack`Private` and `Private` is 9 characters long, it works: > currentFileName is equal to "TestPack.m". However I find this solution > quite cumbersome, is there a more obvious way to do that? > > Thank you > > Geronimo > > >