Question

我想在单击按钮时执行长任务。我希望此任务阻止UI，因为在任务完成之前应用程序无法运行。但是，我想向用户表明发生了一些事情，所以我有一个BusyIndicator（在渲染线程上运行）并设置为在操作开始之前显示。但是，它永远不会呈现。为什么呢？

main.cpp中：

#include <QGuiApplication>
#include <QQmlApplicationEngine>
#include <QQmlContext>
#include <QDateTime>
#include <QDebug>

class Task : public QObject
{
    Q_OBJECT
    Q_PROPERTY(bool running READ running NOTIFY runningChanged)

public:
    Task() : mRunning(false) {}

    Q_INVOKABLE void run() {
        qDebug() << "setting running property to true";
        mRunning = true;
        emit runningChanged();

        // Try to ensure that the scene graph has time to begin the busy indicator
        // animation on the render thread.
        Q_ASSERT(QMetaObject::invokeMethod(this, "doRun", Qt::QueuedConnection));
    }

    bool running() const {
        return mRunning;
    }

signals:
    void runningChanged();

private:
    Q_INVOKABLE void doRun() {
        qDebug() << "beginning long, blocking operation";
        QDateTime start = QDateTime::currentDateTime();
        while (start.secsTo(QDateTime::currentDateTime()) < 2) {
            // Wait...
        }
        qDebug() << "finished long, blocking operation";

        qDebug() << "setting running property to false";
        mRunning = false;
        emit runningChanged();
    }

    bool mRunning;
};

int main(int argc, char *argv[])
{
    QGuiApplication app(argc, argv);

    QQmlApplicationEngine engine;
    Task task;
    engine.rootContext()->setContextProperty("task", &task);
    engine.load(QUrl(QStringLiteral("qrc:/main.qml")));

    return app.exec();
}

#include "main.moc"

main.qml：

import QtQuick 2.6
import QtQuick.Window 2.2
import Qt.labs.controls 1.0

Window {
    width: 600
    height: 400
    visible: true

    Shortcut {
        sequence: "Ctrl+Q"
        onActivated: Qt.quit()
    }

    Column {
        anchors.centerIn: parent
        spacing: 20

        Button {
            text: task.running ? "Running task" : "Run task"
            onClicked: task.run()
        }
        BusyIndicator {
            anchors.horizontalCenter: parent.horizontalCenter
            running: task.running
            onRunningChanged: print("BusyIndicator running =", running)
        }
    }
}

调试输出在事件顺序方面看起来是正确的：

setting running property to true
qml: BusyIndicator running = true
beginning long, blocking operation
finished long, blocking operation
setting running property to false
qml: BusyIndicator running = false

Answer 1

QML中的大多数动画都依赖于主线程中管理的属性，因此在主UI线程被阻止时会被阻止。查看http://doc.qt.io/qt-5/qml-qtquick-animator.html可以在主线程被阻止时运行的动画。如果可能的话，我会将操作转移到另一个线程中，这样做更简单，并且允许例如从UI取消操作。

Answer 2

使用Qt::QueuedConnection调用函数并不能保证BusyIndicator有机会开始制作动画。它只是guarantees that：

当控制返回到接收者线程的事件循环时，调用该槽。

可能听起来很有希望的

Another solution是QTimer::singleShot()：

QTimer::singleShot(0, this, SLOT(doRun()));

作为一种特殊情况，一旦处理了窗口系统事件队列中的所有事件，超时为0的QTimer将超时。这可以用来做繁重的工作，同时提供一个活泼的用户界面[...]

然而，这也没有奏效。我不确定为什么。可能是，在内部，渲染/动画不是通过排队调用完成的，因此超时发生得太早。

您可以指定等待的任意时间量：

QTimer::singleShot(10, this, SLOT(doRun()));

这会起作用，但它并不那么好;它只是在猜测。

你需要的是reliable way知道场景图何时开始动画。

<强> main.cpp中：

#include <QGuiApplication>
#include <QQmlApplicationEngine>
#include <QQmlContext>
#include <QDateTime>
#include <QDebug>
#include <QTimer>
#include <QQuickWindow>

class Task : public QObject
{
    Q_OBJECT
    Q_PROPERTY(bool running READ running NOTIFY runningChanged)

public:
    Task(QObject *parent = 0) :
        QObject(parent),
        mRunning(false) {
    }

signals:
    void runningChanged();

public slots:
    void run() {
        qDebug() << "setting running property to true";
        mRunning = true;
        emit runningChanged();

        QQmlApplicationEngine *engine = qobject_cast<QQmlApplicationEngine*>(parent());
        QQuickWindow *window = qobject_cast<QQuickWindow*>(engine->rootObjects().first());
        connect(window, SIGNAL(afterSynchronizing()), this, SLOT(doRun()));
    }

    bool running() const {
        return mRunning;
    }

private slots:
    void doRun() {
        qDebug() << "beginning long, blocking operation";
        QDateTime start = QDateTime::currentDateTime();
        while (start.secsTo(QDateTime::currentDateTime()) < 2) {
            // Wait...
        }
        qDebug() << "finished long, blocking operation";

        QQmlApplicationEngine *engine = qobject_cast<QQmlApplicationEngine*>(parent());
        QQuickWindow *window = qobject_cast<QQuickWindow*>(engine->rootObjects().first());
        disconnect(window, SIGNAL(afterSynchronizing()), this, SLOT(doRun()));

        qDebug() << "setting running property to false";
        mRunning = false;
        emit runningChanged();
    }

private:
    bool mRunning;
};

int main(int argc, char *argv[])
{
    QGuiApplication app(argc, argv);

    QQmlApplicationEngine engine;
    Task task(&engine);
    engine.rootContext()->setContextProperty("task", &task);
    engine.load(QUrl(QStringLiteral("qrc:/main.qml")));

    return app.exec();
}

#include "main.moc"

<强> main.qml ：

import QtQuick 2.6
import QtQuick.Window 2.2
import Qt.labs.controls 1.0

Window {
    width: 600
    height: 400
    visible: true

    Shortcut {
        sequence: "Ctrl+Q"
        onActivated: Qt.quit()
    }

    Column {
        anchors.centerIn: parent
        spacing: 20

        Button {
            text: task.running ? "Running task" : "Run task"
            onClicked: task.run()
        }
        BusyIndicator {
            anchors.horizontalCenter: parent.horizontalCenter
            running: task.running
            onRunningChanged: print("BusyIndicator running =", running)
        }
    }
}

此解决方案依赖于访问应用程序窗口，该窗口 很好，但它消除了任何猜测。请注意，如果我们之后不再断开信号，则每次场景图完成同步时都会继续调用它，因此执行此操作非常重要。

如果您有多个需要此类解决方案的操作，请考虑创建可重用的类：

class BlockingTask : public QObject { Q_OBJECT Q_PROPERTY(bool running READ running NOTIFY runningChanged) public: BlockingTask(QQmlApplicationEngine *engine) : mEngine(engine), mRunning(false) { } bool running() const { return mRunning; } signals: void runningChanged(); public slots: void run() { qDebug() << "setting running property to true"; mRunning = true; emit runningChanged(); QQuickWindow *window = qobject_cast<QQuickWindow*>(mEngine->rootObjects().first()); connect(window, SIGNAL(afterSynchronizing()), this, SLOT(doRun())); } protected: virtual void execute() = 0; private slots: void doRun() { QQuickWindow *window = qobject_cast<QQuickWindow*>(mEngine->rootObjects().first()); disconnect(window, SIGNAL(afterSynchronizing()), this, SLOT(doRun())); execute(); qDebug() << "setting running property to false"; mRunning = false; emit runningChanged(); } private: QQmlApplicationEngine *mEngine; bool mRunning; };

然后，子类只需要担心它们的逻辑：

class Task : public BlockingTask { Q_OBJECT public: Task(QQmlApplicationEngine *engine) : BlockingTask(engine) { } protected: void execute() Q_DECL_OVERRIDE { qDebug() << "beginning long, blocking operation"; QDateTime start = QDateTime::currentDateTime(); while (start.secsTo(QDateTime::currentDateTime()) < 2) { // Wait... } qDebug() << "finished long, blocking operation"; } };

在Qt Quick中阻止操作

2 个答案: