PyQt5,点击菜单并打开新窗口

时间:2017-08-15 08:10:47

标签: python qt pyqt pyqt5 qt-designer

我使用Qt设计器创建了两个不同的窗口,input_window.ui和help_window.ui。这是用于显示输入窗口的python脚本。在输入窗口中,有一个菜单栏(“关于>>帮助”)。单击“帮助”时如何弹出help_window?

  

这是 init .py

import sys
from input_window import Ui_MainWindow
from PyQt5.QtWidgets import QMainWindow, QApplication
from help_window import Ui_Help



class MainWindow(QMainWindow, Ui_MainWindow):
    def __init__(self, parent=None):
        super(MainWindow, self).__init__(parent)
        self.setupUi(self)
        self.Pophelp.triggered.connect(self.Ui_Help)


    def help_window(self):
        self.window=Ui_Help()
        self.window.show()

if __name__ == "__main__":
    app = QApplication(sys.argv)
    window = MainWindow()
    window.show()
    sys.exit(app.exec_())

以下是Ui_Help的代码

from PyQt5 import QtCore, QtGui, QtWidgets

class Ui_Help(object):
    def setupUi(self, Help):
        Help.setObjectName("Help")
        Help.resize(251, 99)
        icon = QtGui.QIcon()
        icon.addPixmap(QtGui.QPixmap("logo.png"), QtGui.QIcon.Normal, QtGui.QIcon.Off)
        Help.setWindowIcon(icon)
        self.gridLayoutWidget = QtWidgets.QWidget(Help)
        self.gridLayoutWidget.setGeometry(QtCore.QRect(9, 9, 231, 81))
        self.gridLayoutWidget.setObjectName("gridLayoutWidget")
        self.gridLayout = QtWidgets.QGridLayout(self.gridLayoutWidget)
        self.gridLayout.setContentsMargins(0, 0, 0, 0)
        self.gridLayout.setObjectName("gridLayout")
        self.plainTextEdit = QtWidgets.QPlainTextEdit(self.gridLayoutWidget)
        font = QtGui.QFont()
        font.setFamily("Times New Roman")
        font.setPointSize(10)
        font.setBold(True)
        font.setWeight(75)
        self.plainTextEdit.setFont(font)
        self.plainTextEdit.setFrameShape(QtWidgets.QFrame.WinPanel)
        self.plainTextEdit.setFrameShadow(QtWidgets.QFrame.Sunken)
        self.plainTextEdit.setLineWidth(1)
        self.plainTextEdit.setSizeAdjustPolicy(QtWidgets.QAbstractScrollArea.AdjustIgnored)
        self.plainTextEdit.setReadOnly(True)
        self.plainTextEdit.setObjectName("plainTextEdit")
        self.gridLayout.addWidget(self.plainTextEdit, 0, 0, 1, 1)

        self.retranslateUi(Help)
        QtCore.QMetaObject.connectSlotsByName(Help)

3 个答案:

答案 0 :(得分:1)

Qt Designer用于以简单的方式实现视图,因此生成的类面向视图,我们的工作是像Ui_MainWindowMainWindow一样实现逻辑,与Ui_Help类似。在你的情况下,我建议你在构建help_window.ui时使用Dialog模板,但是如果选择Widget模板没有问题,两者都非常兼容。

一个简单的解决方案是创建QDialog并在其中实施Ui_Help视图,如下所示:

class MainWindow(QMainWindow, Ui_MainWindow):
    def __init__(self, parent=None):
        super(MainWindow, self).__init__(parent)
        self.setupUi(self)
        self.Pophelp.triggered.connect(self.help_window)

    def help_window(self):
        # If you pass a parent (self) will block the Main Window,
        # and if you do not pass both will be independent,
        # I recommend you try both cases.
        widget = QDialog(self)
        ui=Ui_Help()
        ui.setupUi(widget)
        widget.exec_()

如果在Ui_Help中你想要实现某些逻辑,我建议创建一个类似于MainWindow的类,如下所示:

class Help(QDialog, Ui_Help):
    def __init__(self, parent=None):
        super(Help, self).__init__(parent)
        self.setupUi(self)

class MainWindow(QMainWindow, Ui_MainWindow):
    def __init__(self, parent=None):
        super(MainWindow, self).__init__(parent)
        self.setupUi(self)
        self.Pophelp.triggered.connect(self.help_window)

    def help_window(self):
        widget = Help()
        widget.exec_()

答案 1 :(得分:0)

你没有包含input_window.ui所以很难完全复制你正在做的事情,但我认为你遇到的主要问题来自于这一行:

self.Pophelp.triggered.connect(self.Ui_Help)

您不想将按钮连接到Ui_Help,而是希望将其连接到self.help_window

答案 2 :(得分:0)

如果将help_window更改为以下代码,则有效。

def help_window(self):
    dialog=QtWidgets.QDialog()
    dialog.ui=Ui_Help()
    dialog.ui.setupUi(dialog)
    dialog.exec_()
    dialog.show()