On Jul 1, 8:36 am, Tom Chang <[email protected]> wrote: > web2py experts, > > a newbie here trying to learn web2py on my first app. > > Basically, I want a user to upload a text file (work perfectly right > from web2py). Then, the backend (controller) will analyze the text > files, and then display the output back to the user. > > The problem I ran into is after I retrive the new filename > information, I need to have the controller to open up the text file > and then analyze it. I ran into the following exception: > > f = open(myFile,'r').read() > IOError: [Errno 2] No such file or directory: '' > > The file path is: > /applications/IOS_Audit/uploads/config.file.924bea965800dbdf. > 736172697373612e6a73.js > > So wondering my path is wrong probably? any suggestion on the right > path to use?
It looks like you are using the <fieldname>_newfilename for this, but just to be sure, see http://web2py.com/book/default/section/7/2?search=newfilename The next thing clear is your path: It is relative to your web2py installation (notice it _begins_ with applications). Try joining this with request.env.web2py_path (you can peruse request, among other places, here: http://web2py.com/examples/simple_examples/hello6, and in the same application / example in your own development copy). The generic view in views/generic.html has the code for showing the request, so you can simply craft up an example controller for yourself without a corresponding default view, and the generic one will be used. Hope this helps. Regards, - Yarko > > best regards, > TC
