Pyside Qt 设计器输出文件连接插槽问题答案

【问题标题】：Pyside Qt designer output file connecting slots issuePyside Qt 设计器输出文件连接插槽问题
【发布时间】：2017-06-07 09:49:56
【问题描述】：

我使用 QT Designer 创建了一个对话框，并使用 pyside-uic 转换为 Python

该程序是一个简单的按钮，因为我目前正在学习 pyside，我想将按钮连接到一个名为“buttonpushed”的函数，只是为了在单击按钮时打印“button pushed”但是尽管遵循在线教程我的按钮没有工作。

当我只是在按钮中编写代码时不会发生此错误，但是当我使用设计器并且它变成一个对话框时，我无法让它工作。

from PySide import QtCore, QtGui

class Ui_Dialog(object):
def setupUi(self, Dialog):
    Dialog.setObjectName("Dialog")
    Dialog.resize(400, 300)
    self.pushButton = QtGui.QPushButton(Dialog)
    self.pushButton.setGeometry(QtCore.QRect(160, 120, 71, 23))
    self.pushButton.setObjectName("pushButton")

    self.retranslateUi(Dialog)

    QtCore.QMetaObject.connectSlotsByName(Dialog)


def retranslateUi(self, Dialog):
    Dialog.setWindowTitle(QtGui.QApplication.translate("Dialog", "Dialog", None, QtGui.QApplication.UnicodeUTF8))
    self.pushButton.setText(QtGui.QApplication.translate("Dialog", "PushButton", None, QtGui.QApplication.UnicodeUTF8))


def buttonpushed(self):
    print 'buttonpushed'


if __name__ == "__main__":
import sys
app = QtGui.QApplication(sys.argv)
Dialog = QtGui.QDialog()
ui = Ui_Dialog()
ui.setupUi(Dialog)
Dialog.show()
sys.exit(app.exec_())

【问题讨论】：

标签： python pyqt pyqt4 pyside pyqt5

【解决方案1】：

你要做的就是把信号和想要的功能连接起来，你可以在课堂内外做。

{yourobject}.{object signal}.connect({your slot})

代码：

from PySide import QtCore, QtGui

class Ui_Dialog(object):
    def setupUi(self, Dialog):
        Dialog.setObjectName("Dialog")
        Dialog.resize(400, 300)
        self.pushButton = QtGui.QPushButton(Dialog)
        self.pushButton.setGeometry(QtCore.QRect(160, 120, 71, 23))
        self.pushButton.setObjectName("pushButton")

        self.retranslateUi(Dialog)

        QtCore.QMetaObject.connectSlotsByName(Dialog)

        self.pushButton.clicked.connect(self.buttonpushed)



    def retranslateUi(self, Dialog):
        Dialog.setWindowTitle(QtGui.QApplication.translate("Dialog", "Dialog", None, QtGui.QApplication.UnicodeUTF8))
        self.pushButton.setText(QtGui.QApplication.translate("Dialog", "PushButton", None, QtGui.QApplication.UnicodeUTF8))


    def buttonpushed(self):
        print('buttonpushed')


if __name__ == "__main__":
    import sys
    app = QtGui.QApplication(sys.argv)
    Dialog = QtGui.QDialog()
    ui = Ui_Dialog()
    ui.setupUi(Dialog)
    Dialog.show()
    sys.exit(app.exec_())

另一种解决方案：

from PySide import QtCore, QtGui

class Ui_Dialog(object):
    def setupUi(self, Dialog):
        Dialog.setObjectName("Dialog")
        Dialog.resize(400, 300)
        self.pushButton = QtGui.QPushButton(Dialog)
        self.pushButton.setGeometry(QtCore.QRect(160, 120, 71, 23))
        self.pushButton.setObjectName("pushButton")

        self.retranslateUi(Dialog)

        QtCore.QMetaObject.connectSlotsByName(Dialog)



    def retranslateUi(self, Dialog):
        Dialog.setWindowTitle(QtGui.QApplication.translate("Dialog", "Dialog", None, QtGui.QApplication.UnicodeUTF8))
        self.pushButton.setText(QtGui.QApplication.translate("Dialog", "PushButton", None, QtGui.QApplication.UnicodeUTF8))


    def buttonpushed(self):
        print('buttonpushed')


if __name__ == "__main__":
    import sys
    app = QtGui.QApplication(sys.argv)
    Dialog = QtGui.QDialog()
    ui = Ui_Dialog()
    ui.setupUi(Dialog)
    ui.pushButton.clicked.connect(ui.buttonpushed)
    Dialog.show()
    sys.exit(app.exec_())

【讨论】：

太好了，效果很好，有什么方法可以从 Ui_Dialog 类外部连接它吗？
@WilliamHamilton 更新我的答案，如果它可以帮助您将其标记为正确，请。