擅长:python、mysql、java
<p>看起来URL示例不好。尝试www.google.com。我已经为PyQt5修改了你的代码。它起作用了。在</p>
<pre><code> from PyQt5.QtCore import QUrl
from PyQt5.QtWidgets import QApplication, QMainWindow
from PyQt5.QtNetwork import QNetworkAccessManager, QNetworkRequest
import sys
def printContent():
answerAsText = bytes(replyObject.readAll()).decode("utf-8")
print(answerAsText)
class mainClass():
def my_exception_hook(exctype, value, traceback):
print(exctype, value, traceback)
sys._excepthook(exctype, value, traceback)
sys.exit(1)
sys.excepthook = my_exception_hook
if __name__ == '__main__':
app = QApplication(sys.argv)
url = QUrl("http://www.google.com")
request = QNetworkRequest()
request.setUrl(url)
manager = QNetworkAccessManager()
replyObject = manager.get(request)
replyObject.finished.connect(printContent)
sys.exit(app.exec_())
</code></pre>