Question

I have several large zip file that contain a dir structure that I must maintain. Currently to unzip them I am using

    zip = zipfile.ZipFile(self.fileName)        
    zip.extractall(self.destination)
    zip.close()

The problem is that these process can take upwards of 3-5 minutes and I have no feedback that they are still working. What I would like to do is output the name of the file currently being unziped to the status bar of my gui. What I have in mind is something like

    zip = zipfile.ZipFile(self.fileName)
    zipNameList = zipfile.namelist(self.fileName)
    for item in zipNameList:
        self.SetStatusText("Unzipping" + str(item))
        zip.extract(item)
    zip.close()

The problem with this is that it does not create the correct dir structure. I am not sure that this is even the best way to go about it.

I was also looking into using wx.progressdialog but could not come up with a way to have it show progress of the zip.extractall(filename).

Answer 1

I got it to an acceptable solution – Though I think I would prefer it thread it eventually.

def unzipItem(self, fileName, destination)
    print "--unzipItem--"
    zip = zipfile.ZipFile(fileName)
    nameList = zip.namelist()

    #get the amount of files in the file to properly size the progress bar
    fileCount = 0
    for item in nameList:
        fileCount += 1

    #Built progress dialog
    dlg = wx.ProgressDialog("Unziping files",
                           "An informative message",
                           fileCount,
                           parent = self,
                           )

    keepGoing = True
    count = 0

    for item in nameList:
        count += 1
        dir,file = os.path.split(item)
        print "unzip " + file

        #update status bar
        self.SetStatusText("Unziping " + str(item))
        #update progress dialog
        (keepGoing, skip) = dlg.Update(count, file)
        zip.extract(item,destination)

    zip.close()