Question

我正在尝试从工作线程显示消息对话框。使用插槽和信号是与QML对象通信/调用QML对象的标准方式，但是当对话框出现时，它的按钮无法点击/无响应。

的main.cpp

tester* test = new tester;
QtConcurrent::run(test,tester::testFunction);

tester.cpp

#include "tester.h"
#include <QQmlEngine>
#include <QQmlComponent>

tester::tester(QObject *parent) : QObject(parent) {
    QObject::connect(this,SIGNAL(show()),this,SLOT(showSlot()));
}
void tester::testFunction() {
    emit show();
}
void tester::showSlot(){
    QQmlEngine engine;
    QQmlComponent component(&engine, QUrl(QLatin1String("qrc:/BlockingDialog.qml")));
    QObject *object = component.create();
    QMetaObject::invokeMethod(object, "open");
}

tester.h

#include <QObject>

class tester : public QObject{
    Q_OBJECT
public:
    explicit tester(QObject *parent = 0);
    void testFunction();
signals:
    void show();
public slots:
    void showSlot();
};

BlockingDialog.qml

import QtQuick 2.7
import QtQuick.Dialogs 1.2

MessageDialog {
    id:dialog    
}

Answer 1

您正在showSlot()中在堆栈上创建QML引擎，因此在函数完成时它将被销毁。加载QML文件的典型方法是在main()。

中创建堆栈上的QML引擎

从Slot qt创建时，MessageDialog会冻结

1 个答案: