问题
The title says pretty much everything.
Lets say I have this simple application:
main.py >>>
import sys
from PyQt5.QtCore import QUrl
from PyQt5.QtWidgets import QApplication
from PyQt5.QtQuick import QQuickView
# Main Function
if __name__ == '__main__':
# Create main app
myApp = QApplication(sys.argv)
# Create a label and set its properties
appLabel = QQuickView()
appLabel.setSource(QUrl('main.qml'))
# Show the Label
appLabel.show()
# Execute the Application and Exit
myApp.exec_()
sys.exit()
main.qml >>>
import QtQuick 2.0
Rectangle {
width: 250; height: 175
Text {
id: helloText
anchors.verticalCenter: parent.verticalCenter
anchors.horizontalCenter: parent.horizontalCenter
text: "Hello World!!!\n Traditional first app using PyQt5"
horizontalAlignment: Text.AlignHCenter
}
}
Now this example is working fine. But lets say I make a typo in main.qml and I write heigth instead of height. Then the python code will work just fine but it will launch an empty window without any error message.
What shall I do to see errors from .qml file in my python console? Finding typo in 6000 lines of code is extremely painful.
I am using PyQt 5.5.1, Anaconda 2.4.1 (Python 3.5.1), Windows 8.1
回答1:
If all you want is to see error output on the console, you don't need to do anything, because Qt automatically does that anyway. For example, if I change height
to heigth
in your example, the following message is printed on stderr:
file:///home/foo/test/main.qml:4:17: Cannot assign to non-existent property "heigth" width: 250; heigth: 175
If you want to raise an exception within your application, you can connect to the statusChanged signal and get the details from the errors method:
def handleStatusChange(status):
if status == QQuickView.Error:
errors = appLabel.errors()
if errors:
raise Exception(errors[0].description())
appLabel = QQuickView()
appLabel.statusChanged.connect(handleStatusChange)
来源:https://stackoverflow.com/questions/35384677/pyqt-qml-error-console-missing