无法启动设计的 pyQt 应用程序 [英] Can't start designed pyQt application

问题描述

我刚刚在 pyQt Designer 5 中设计了我的应用程序，将 main.ui 生成为 main.py，将 assets.qrc 生成为 assets_rc.py.没有错误，当我现在从终端运行 main.py 时，什么也没有发生.我错过了一步吗?我现在应该编辑我的 main.py 文件吗?

I have just designed my application inside pyQt Designer 5, generated my main.ui to main.py and my assets.qrc to assets_rc.py. No errors, when I now run main.py from my terminal nothing happends. Have I missed a step? Am I supposed to edit my main.py file now?

干杯！

Python 3.3.0pyQT 5

Python 3.3.0 pyQT 5

推荐答案

这适用于 PyQt4，但应该与 PyQt5 相同.

This is for PyQt4, but should be the same for PyQt5.

假设您的用户界面名为mainwindow.ui".使用 pyuic4 将其编译为mainWindowUi.py"(或其他任何内容，只需保留名称即可).

Let's say your ui is called "mainwindow.ui". Compile this with pyuic4 into "mainWindowUi.py" (or whatever, just stick to the name).

现在创建一个包含或多或少此内容的文件mainWindow.py":

Now create a file "mainWindow.py" with more or less this content:

from PyQt4 import QtGui
from mainWindowUi import Ui_MainWindow #same name as appears in mainWindowUi.py

class MainWindow (QtGui.QMainWindow): #Or wherever you are inheriting from
    def __init__ (self, parent = None):
        super (MainWindow, self).__init__ ()
        self.ui = Ui_MainWindow () #same name as appears in mainWindowUi.py
        self.ui.setupUi (self)

    #implement slots and signals and other funny things

现在创建一个包含或多或少此内容的文件program.py":

Now create a file "program.py" with more or less this content:

#! /usr/bin/python3.3

import sys
from PyQt4 import QtGui
from mainWindow import MainWindow

def main():
    app = QtGui.QApplication (sys.argv)
    m = MainWindow ()
    m.show ()
    sys.exit (app.exec_ () )


if __name__ == '__main__':
    main ()

运行 program.py 文件.这或多或少是 Qt 应用程序的骨架.

Run the program.py file. This is more or less the skeleton of a Qt application.

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