警告消息“无匹配信号”

时间:2018-12-21 07:07:36

标签: c++ qt qt4 ros vlc-qt

我正在基于qt-ros使用qt4来构建应用程序。

但是有一个问题,signal & slot不起作用。

我正在使用的vlc-qt库提供了一个称为play的信号功能,如下面的链接所示。 vlc-qt

我尝试通过创建适当的QMetaObject :: connectSlotsByName函数来连接到slot方法,但不适用于警告“无匹配信号”。

在mainWindow.h

public Q_SLOTS:
    void on_vListPlayer_played();

和在mainWindow.cpp

void MainWindow::on_vListPlayer_played()
{
    ROS_INFO("player started!------------------------------");
}
...
MainWindow::MainWindow(int argc, char** argv, QWidget *parent)
: QMainWindow(parent)
, qnode(argc,argv)
{
    ui.setupUi(this); // Calling this incidentally connects all ui's triggers to on_...() callbacks in this class.

    // UI Init
    QWidget* mainWidget = new QWidget(this);
    this->setCentralWidget(mainWidget);
    mainWidget->setStyleSheet("background-color: black;");
    QVBoxLayout* mainLayout = new QVBoxLayout;
    mainLayout->setMargin(0);
    mainLayout->setSpacing(0);
    mainWidget->setLayout(mainLayout);

    m_vVideoWidget = new VlcWidgetVideo;
    mainLayout->addWidget(m_vVideoWidget);

    m_vInstance = new VlcInstance(VlcCommon::args(), this);
    m_vPlayer = new VlcMediaPlayer(m_vInstance);
    m_vPlayer->setVideoWidget(m_vVideoWidget);

    vListPlayer = new VlcMediaListPlayer(m_vPlayer, m_vInstance);
    QObject::connect(vListPlayer, SIGNAL(played()), this, SLOT(on_vListPlayer_played()));

    m_vVideoWidget->setMediaPlayer(m_vPlayer);

    m_vList = new VlcMediaList(m_vInstance);
    openVideoes(m_DataPath);

    vListPlayer->setMediaList(m_vList);
    vListPlayer->setPlaybackMode(Vlc::PlaybackMode::Repeat);

    vListPlayer->mediaPlayer()->play();
...
}

在MediaListPlayer.h中(vlc-qt lib)

class VLCQT_CORE_EXPORT VlcMediaListPlayer : public QObject
{
    Q_OBJECT
......
public Q_SLOTS:
    void itemAt(int index);
    void next();
    void play();
    void previous();
    void stop();

Q_SIGNALS:

    void played();
    void nextItemSet(VlcMedia *media);
    void nextItemSet(libvlc_media_t *media);
    void stopped();

1 个答案:

答案 0 :(得分:1)

您使用的是Qt Designer,生成的代码(由ui.setupUi(this);调用)会调用QMetaObject::connectSlotsByName(QObject *object)

根据Qt documentation,这会尝试连接名称与以下模式匹配的所有插槽:void on_<object name>_<signal name>(<signal parameters>);

由于插槽void on_vListPlayer_played()与模式匹配,因此尝试进行连接。但是失败,因为您没有任何对象named vListPlayer

在您的情况下,我建议您重命名插槽,以使其与模式不匹配并且不会自动连接。