PyQt 打开文件对话框 - 显示路径名

ola*_*nre 2 user-interface file-upload file pyqt python-2.7

使用 PyQt,我创建了一个按钮,可以从任何目录中选择和上传文件。如何在按钮旁边创建一个文本框来显示所选或打开的文件的路径?

这是我的示例代码:

    self.uploadButton = QtGui.QPushButton('UPLOAD SDF', self)
    self.runfilterButton = QtGui.QPushButton('Run Filter', self)
    self.printimagesButton = QtGui.QPushButton('Display Matches Images', self)
    self.listmatchesButton = QtGui.QPushButton('List Matches', self)

    self.uploadButton.move (100, 50)
    self.runfilterButton.move (400,50)
    self.printimagesButton.move (200, 100)
    self.printimagesButton.resize (200, 50)
    self.listmatchesButton.move (200, 150)
    self.listmatchesButton.resize (200, 50)
    hBoxLayout = QtGui.QHBoxLayout()

    hBoxLayout.addWidget(self.uploadButton)
    hBoxLayout.addWidget(self.runfilterButton)
    self.setLayout(hBoxLayout)
    # Signal Init.
    self.connect(self.uploadButton, QtCore.SIGNAL('clicked()'), self.open)
    self.runfilterButton.clicked.connect(runfilterx)
    self.printimagesButton.clicked.connect(printimages)
    self.listmatchesButton.clicked.connect(listmatches)
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Dig*_*yad 6

首先,您需要创建一个 TextEdit 对象,例如self.myTextBox = QtGui.QTextEdit(self). 之后,将其放置在屏幕上您想要的任何位置。然后你需要将它与你的open函数连接起来。它应该看起来像这样。

def open(self):
    fileName = QtGui.QFileDialog.getOpenFileName(self, 'OpenFile')
    self.myTextBox.setText(fileName)
    print(fileName)
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这将在您的textBox.