CODEWITHSUNDEEP
pythonpyqt
Jacky Wu
Jacky Wu

Reputation: 27

How to get the file name and show it on Qlistwidget?

I want to make a song list by open the QfileDialog to add the file .
Now I can play the song of list , When I click the item of QlistWidget.
But the item name is its path .
I want to show the file name on QlistWidget .
And when I click the item of QlistWidget , it has to transfer the path to openaudio() method .

Here is part of code :

expand='Image Files(*.mp3 *.wav)'
tips=open the file''
path = QtGui.QFileDialog.getOpenFileName(self, tips,QtGui.QDesktopServices.storageLocation(QtGui.QDesktopServices.MusicLocation), expand)
if path:
   mlist.append(path)
   index=mlist.index(path) 
   self.ui.listWidget.insertItem(index,mlist[index])

Here is code of openaudio() :

def openaudio(self,path):
        self.connect(self.ui.listWidget,QtCore.SIGNAL('currentTextChanged(QString)'),self.ui.label_4,QtCore.SLOT('setText(QString)'))
        index=self.ui.listWidget.currentRow()        
        path=mlist[index]

        self.mediaObject.setCurrentSource(phonon.Phonon.MediaSource(path))
        self.mediaObject.play()

By the way ,how can I open multiple files at once ?

Upvotes: 0

Views: 1534

Answers (1)

sonovice
sonovice

Reputation: 831

One way would be subclassing QListWidgetItem:

class MyItem(QListWidgetItem):
    def __init__(self, path, parent=None):
        self.path = path

        import os
        filename = os.path.basename(self.path)
        super().__init__(filename)

And then connecting your QListWidget to your openaudio(path) method:

self.ui.listWidget.itemClicked.connect(lambda n: openaudio(n.path))

Apart from that specific question your code seems to have some other issues. Using an additional list (mlist) is not needed in this particular case:

path = QtGui.QFileDialog.getOpenFileName(self, tips, QtGui.QDesktopServices.storageLocation(QtGui.QDesktopServices.MusicLocation), expand)
if path:
    self.ui.listWidget.addItem(MyItem(path))

and

def openaudio(self, path):
    # Do not do this here! Connections should usually be made in the init() method of the container!
    # self.connect(self.ui.listWidget,QtCore.SIGNAL('currentTextChanged(QString)'),self.ui.label_4,QtCore.SLOT('setText(QString)'))

    # Also take a look at the 'new' PyQt signals & slots style:
    # self.ui.listWidget.currentTextChanged.connect(self.ui.label_4.setText)

    self.mediaObject.setCurrentSource(phonon.Phonon.MediaSource(path))
    self.mediaObject.play()

Upvotes: 2

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