Question

在我的应用程序中，我使用QtLocation来显示地图。由于只有QML API来渲染地图，因此这是我的QML文件：

import QtQuick 2.0
import QtPositioning 5.5
import QtLocation 5.5

Item{
    anchors.fill: parent

    Plugin{
        id: osmplugin
        name: "osm"
    }

    Map {
        anchors.fill: parent
        id: map
        plugin: osmplugin;
        zoomLevel: (maximumZoomLevel - minimumZoomLevel)/2
        center {
            // The Qt Company in Oslo
            latitude: 59.9485
            longitude: 10.7686
        }
    }

    function bbox(){
        return map.visibleRegion;
    }
}

在C ++代码中，我需要知道地图窗口小部件中当前可见的区域，QML Map具有属性visibleRegion http://doc.qt.io/qt-5/qml-qtlocation-map.html#visibleRegion-prop

但是我不明白如何从C ++代码中获取它，因为QGeoShape是抽象的;

我试过了：

    QQuickItem* map = mMap->rootObject();
    QGeoRectangle rect;
    bool ok = QMetaObject::invokeMethod( map, "bbox",  Qt::DirectConnection, Q_RETURN_ARG( QGeoRectangle, rect ) );
    if ( !ok )
        qDebug() << " Shit happens!";

    qDebug() << rect.isValid();

但它没有帮助。请告诉我如何从QML Map获得可见的矩形。

Answer 1

正确的语法是：

QQuickItem* map = mMap->rootObject();
QVariant ret;
bool ok = QMetaObject::invokeMethod( map, "bbox",  Qt::DirectConnection, Q_RETURN_ARG( QVariant, ret ) );
if ( !ok ){
    qWarning( "Fail to call qml method" );
}
QGeoRectangle rect = qvariant_cast<QGeoRectangle>( ret );
mNorth = rect.topLeft().latitude();
mSouth = rect.bottomLeft().latitude();
mWest  = rect.topLeft().longitude();
mEast  = rect.topRight().longitude();

QML地图可见区域

1 个答案: