Qt Quick 中的阻塞操作

Question

我想在单击按钮时执行一项较长的任务。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，我想向用户表明正在发生某些事情，因此我有一个 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？

main.cpp:

#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

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

The slot is invoked when control returns to the event loop of the receiver's thread.

Another solution 听起来很有希望的是 QTimer::singleShot():

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

As a special case, a QTimer with a timeout of 0 will time out as soon as all the events in the window system's event queue have been processed. This can be used to do heavy work while providing a snappy user interface [...]

但是，这也行不通。我不确定为什么。可能是在内部，rendering/animation 不是通过排队调用完成的，因此超时发生得太早了。

您可以指定任意的等待时间：

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

这会起作用，但不是很好；这只是猜测。

您需要的是 reliable way 知道场景图何时开始动画。

main.cpp:

#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"

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)
        }
    }
}

此解决方案依赖于对应用程序 window 的访问权限，这并不好，但它消除了任何猜测。请注意，如果我们之后不断开信号，则每次场景图完成同步时都会继续调用它，因此执行此操作很重要。

如果您有多个操作需要此类解决方案，请考虑创建一个可重复使用的 class:

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;
};

那么子class们只需要担心他们的逻辑：

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";
    }
};

Answer 2

QML 中的大多数动画都依赖于主线程中管理的属性，因此当主 UI 线程被阻塞时也会被阻塞。查看 http://doc.qt.io/qt-5/qml-qtquick-animator.html 以了解在主线程被阻塞时可以运行的动画。如果可能的话，我会将操作移到另一个线程中，这样会简单得多，并且还允许例如从 UI.

取消操作

Qt Quick 中的阻塞操作

Blocking operation in Qt Quick

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qml

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