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如何实现PyQt5与QML响应彼此发送的信号

2025-02-12 14:55:14

推荐回答（1个）

回答1：

qml和cpp中的类没有建立关联的话就不会响应信号 QDeclarativeView login; login.rootContext()->setContextProperty("ycmainui",YcMainUi::YcMain()); (YcMainUi::YcMain() cpp中类的单实例) qml中： Connections{ target: ycmainui onTo_update。

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