Hi mdipierro and Yarko, Thanks for the information. It does help and work now!!!
best regards, TC On Jul 1, 11:57 am, Yarko Tymciurak <[email protected]> wrote: > 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, > seehttp://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
