if filename is the file from the database filename='config.file.924bea965800dbdf.736172697373612e6a73.js'
you can use data = open(os.path.join(request.folder,'uploads',filename),'rb').read() On 1 Lug, 08:36, 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? > > best regards, > TC
