python - QWebKit linkClicked 信号永远不会触发

Question

import sys

from PyQt4.QtGui import *
from PyQt4.QtCore import *
from PyQt4.QtWebKit import QWebView

app = QApplication(sys.argv)
web_view = QWebView()
def url_changed(url):  print 'url changed: ', url
def link_clicked(url):  print 'link clicked: ', url
def load_started():  print 'load started'
def load_finished(ok):  print 'load finished, ok: ', ok
web_view.connect(web_view, SIGNAL("urlChanged(const QUrl&)"), url_changed)
web_view.connect(web_view, SIGNAL("linkClicked(const QUrl&)"), link_clicked)
web_view.connect(web_view, SIGNAL('loadStarted()'), load_started)
web_view.connect(web_view, SIGNAL('loadFinished(bool)'), load_finished)
web_view.load(QUrl('http://google.com'))
web_view.show()
sys.exit(app.exec_())

linkClicked 信号不起作用。其他信号有效。Win XP 上的 Qt 4.6.2。

Answer 1

必须适当设置链接委托策略才能发出 linkClicked信号。

import sys

from PyQt4.QtGui import *
from PyQt4.QtCore import *
from PyQt4.QtWebKit import QWebPage, QWebView

app = QApplication(sys.argv)
web_view = QWebView()
web_view.page().setLinkDelegationPolicy(QWebPage.DelegateAllLinks)
def url_changed(url):  print 'url changed: ', url
def link_clicked(url):  print 'link clicked: ', url
def load_started():  print 'load started'
def load_finished(ok):  print 'load finished, ok: ', ok
web_view.connect(web_view, SIGNAL("urlChanged(const QUrl&)"), url_changed)
web_view.connect(web_view, SIGNAL("linkClicked(const QUrl&)"), link_clicked)
web_view.connect(web_view, SIGNAL('loadStarted()'), load_started)
web_view.connect(web_view, SIGNAL('loadFinished(bool)'), load_finished)
web_view.load(QUrl('http://google.com'))
web_view.show()
sys.exit(app.exec_())