python-如果没有父窗口,则无法在PyQt中创建新窗口

我开始用PyQt用Python编写一个简单的文本编辑器,然后遇到了这个问题:对于“新建文档”按钮,我想打开一个新的空文本编辑器,无论第一个窗口发生什么情况,它都保持打开状态.问题是,让我显示窗口的唯一方法是将self作为参数发送(使其成为父项),从而导致在父项关闭时第二个窗口关闭.

这是我的构造函数:

class Main(QtGui.QMainWindow):

def __init__(self, ctrl, parent=None):
    QtGui.QMainWindow.__init__(self, parent)

这是打开新窗口的方法:

def new(self):
    repo = Repository()
    ctrl = Controller(repo)
    new_win = Main(ctrl)
    new_win.show()

注意:如果此处的代码不起作用,则不会显示第二个窗口

编辑:决定我应该发布我所有的代码,所以它去了:

import sys

from PyQt4 import QtGui

from src.controller import Controller
from src.repository import Repository


class Main(QtGui.QMainWindow):

    nr_of_instances = -1

    def __init__(self, ctrl, parent=None):
        QtGui.QMainWindow.__init__(self, parent)
        Main.nr_of_instances += 1
        #ui elements
        self._toolbar = None
        self._menuBar = None
        self._file = None
        self._edit = None
        self._view = None
        self._formatBar = None
        self._statusBar = None
        self._text = None

        #actions
        self._newAction = None
        self._openAction = None
        self._saveAction = None
        self._saveAsAction = None

        #
        self._controller = ctrl

        self.init_ui()

    def init_ui(self):
        self._text = QtGui.QTextEdit(self)
        self.setCentralWidget(self._text)

        self.init_toolbar()
        self.init_formatBar()
        self.init_menuBar()
        self._statusBar = self.statusBar()

        self.setGeometry(50+(50*Main.nr_of_instances), 100+(50*Main.nr_of_instances), 800, 400)

        self.setWindowTitle("KekWriter")

    @staticmethod
    def new(self):
        repo = Repository()
        ctrl = Controller(repo)
        spawn = Main(ctrl)
        spawn.show()

    def set_title(self):
        if self._controller.get_file_name() != 0:
            new_title = self.windowTitle().split(" - ")
            new_title[0].strip()
            self.setWindowTitle(new_title[0]+" - "+self._controller.get_file_name())

    def open(self):
        fn = QtGui.QFileDialog.getOpenFileName(self, 'Open File', ".")
        self._controller.set_file_name(fn)
        try:
            if fn != '':
                self._text.setText(self._controller.open())
            self.set_title()
        except UnicodeDecodeError as msg:
            QtGui.QMessageBox.information(self, "Eroare!", "Tip de fisier invalid!")

    def _save_as(self):
        fn = QtGui.QFileDialog.getSaveFileName(self, 'Save File As', ".")
        print(fn)
        if fn != '':
            self._controller.set_file_name(fn)
            self._controller.save(self._text.toPlainText())
            self.set_title()

    def save(self):
        fn = self._controller.get_file_name()
        if fn == '':
            self._save_as()
        else:
            self._controller.save(self._text.toPlainText())
            self.set_title()

    def init_toolbar(self):
        self._newAction = QtGui.QAction(QtGui.QIcon("icons/new.png"), "New", self)
        self._newAction.setStatusTip("Creates a new document")
        self._newAction.setShortcut("Ctrl+N")
        self._newAction.triggered.connect(self.new)

        self._openAction = QtGui.QAction(QtGui.QIcon("icons/open.png"), "Open", self)
        self._openAction.setStatusTip("Opens existing document")
        self._openAction.setShortcut("Ctrl+O")
        self._openAction.triggered.connect(self.open)

        self._saveAction = QtGui.QAction(QtGui.QIcon("icons/save.png"), "Save", self)
        self._saveAction.setStatusTip("Saves current document")
        self._saveAction.setShortcut("Ctrl+S")
        self._saveAction.triggered.connect(self.save)

        self._saveAsAction = QtGui.QAction(QtGui.QIcon("icons/save_as.png"), "Save as", self)
        self._saveAsAction.setStatusTip("Saves current document with another name")
        self._saveAsAction.setShortcut("Ctrl+Shift+S")
        self._saveAsAction.triggered.connect(self._save_as)

        self._toolbar = self.addToolBar("Options")

        self._toolbar.addAction(self._newAction)
        self._toolbar.addAction(self._openAction)
        self._toolbar.addAction(self._saveAction)
        self._toolbar.addAction(self._saveAsAction)

        self.addToolBarBreak()

    def init_menuBar(self):
        self._menuBar = self.menuBar()
        self._file = self._menuBar.addMenu("File")
        self._edit = self._menuBar.addMenu("Edit")
        self._view = self._menuBar.addMenu("View")

        #file
        self._file.addAction(self._newAction)
        self._file.addAction(self._openAction)
        self._file.addAction(self._saveAction)
        self._file.addAction(self._saveAsAction)

    def init_formatBar(self):
        self._formatBar = self.addToolBar("Format")
        self.addToolBarBreak()


def main():
    app = QtGui.QApplication(sys.argv)
    repo = Repository()
    ctrl = Controller(repo)
    main = Main(ctrl)
    main.show()
    sys.exit(app.exec_())


if __name__ == "__main__":
    main()

解决方法:

您的第二个窗口不显示的原因是该窗口对象超出范围并被垃圾回收.您需要将对第二个窗口的引用存储在某个地方,除非关闭该窗口,否则它不会被删除.

我可以想到几种方法来实现此目的,但实际上如何确定程序结构完全取决于您.

>具有Main的类属性,该属性是一个列表.因此,该列表将对所有实例通用,并且您可以在创建新实例后将该列表追加到该实例中.只要存在一个实例,就不应对窗口进行垃圾收集(我认为)
>最初不要实例化QMainWindow,而是要实例化一个将包含对Windows的引用的类.当创建一个新窗口时,该对象中对新窗口的引用会引起痛处.

希望这可以让您对问题有一个了解,以便以适合您程序布局的方式解决

更新

有关选择方法2的大致指导,请执行以下操作:

class WindowContainer(object):
    def __init__(self):
        self.window_list = []
        self.add_new_window()

    def add_new_window(self):
        repo = Repository()
        ctrl = Controller(repo)
        spawn = Main(ctrl, self)
        self.window_list.append(spawn)
        spawn.show()


class Main(QtGui.QMainWindow):
    def __init__(self, ctrl, window_container, parent=None):
        QtGui.QMainWindow.__init__(self, parent) 
        ...
        self.window_container = window_container
        ...

    ...

    def init_toolbar(self):
        ...
        self._newAction.triggered.connect(self.window_container.add_new_window)
        ...

    ...


def main():
    app = QtGui.QApplication(sys.argv)
    # this variable will never go out of scope
    window_container = WindowContainer()
    sys.exit(app.exec_())
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