Qt Quick中的阻止操作

Question

单击按钮时，我想执行长时间的任务。我想要这个任务来阻止UI，因为应用程序在任务完成之前无法运行。但是，我想向用户指出发生了什么，因此我有一个BusyIndicator（在渲染线程上运行），并且在操作开始之前设置为显示。但是，它从不呈现。为什么？

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) 
     } 
    } 
} 

调试输出看起来是正确的事件顺序方面：

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

用一个函数调用Qt::QueuedConnection不能保证BusyIndicator有机会开始动画。它只是guarantees that：

当控制权返回到接收方线程的事件循环时，将调用此插槽。

Another solution这听起来可能看好的是QTimer::singleShot()：

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

作为一个特殊的情况下，为0的超时QTimer将尽快超时作为窗口系统的事件队列中的所有事件已经被处理。这可以用来做繁重的工作，同时提供一个快速的用户界面[...]

但是，这也行不通。我不知道为什么。这可能是因为内部的渲染/动画不是通过排队调用完成的，所以超时发生得太早。

您可以指定任意的时间量等待：

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

这是可行的，但它不是很好;这只是猜测。

您需要的是知道场景图形何时开始动画的reliable way。

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" 

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) 
     } 
    } 
} 

该解决方案依赖于具有访问应用程序窗口，这是不那很好，但它消除了任何猜测。请注意，如果我们之后没有断开与信号的连接，每次场景图完成同步时都会继续调用它，所以这很重要。

如果你有几个操作，将需要这种类型的解决方案，可以考虑创建一个可重用的类：

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，子类只需要担心自己的逻辑：

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

大多数动画在QML中取决于在主线程中管理的属性，并因此在主UI线程被阻塞时被阻塞。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，并且还允许取消用户界面的操作。