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分享程序员开发的那些事...
且构网 - 分享程序员编程开发的那些事

窗口不会在外部 url 链接点击上打开新窗口或标签

更新时间:2022-03-06 21:51:02

你必须重写 createWindow 方法并返回一个 QWebEngineView,但是要使对象不被干扰,它必须是另一个窗口的子窗口或者是某个窗口的一部分具有更长生命周期的容器.

You have to override the createWindow method and return a QWebEngineView, but for the object not to be distruded it must be the child of another window or be part of a container that has a longer life cycle.

from PyQt5 import QtCore, QtGui, QtWidgets, QtWebEngineWidgets

class WebEnginePage(QtWebEngineWidgets.QWebEnginePage):
    def acceptNavigationRequest(self, url,  _type, isMainFrame):
        if _type == QtWebEngineWidgets.QWebEnginePage.NavigationTypeLinkClicked:
            return True
        return super(WebEnginePage, self).acceptNavigationRequest(url, _type, isMainFrame)

class HtmlView(QtWebEngineWidgets.QWebEngineView):
    def __init__(self, windows, *args, **kwargs):
        super(HtmlView, self).__init__(*args, **kwargs)
        self.setPage(WebEnginePage(self))
        self._windows = windows
        self._windows.append(self)

    def createWindow(self, _type):
        if QtWebEngineWidgets.QWebEnginePage.WebBrowserTab:
            v = HtmlView(self._windows)
            v.resize(640, 480)
            v.show()
            return v

if __name__ == '__main__':
    import sys

    app = QtWidgets.QApplication(sys.argv)
    windows = []
    w = HtmlView(windows)
    w.load(QtCore.QUrl("https://gmail.com"));
    w.show()
    sys.exit(app.exec_())