Openlayers为绘制框和矩形提供了有用的功能,并且还具有ol.geom.Geometry.prototype.rotate(angle,anchor),用于围绕某个锚旋转几何体。是否可以在修改时锁定框/矩形的旋转?
使用位于here的OpenLayers示例绘制具有特定旋转的框以说明要点:
我希望盒子/矩形能够保持其旋转,同时仍然可以将侧面拖得更长更短。有没有一种简单的方法来实现这一目标?
回答我提出的解决方案。
首先,将特征添加到ModifyInteraction,以便您可以通过拖动特征的角来进行修改。
this.modifyInteraction = new Modify({
deleteCondition: eventsCondition.never,
features: this.drawInteraction.features,
insertVertexCondition: eventsCondition.never,
});
this.map.addInteraction(this.modifyInteraction);
此外,在事件“modifystart”和“modifyend”上添加事件处理程序。
this.modifyInteraction.on("modifystart", this.modifyStartFunction);
this.modifyInteraction.on("modifyend", this.modifyEndFunction);
“modifystart”和“modifyend”的功能如下所示。
private modifyStartFunction(event) {
const features = event.features;
const feature = features.getArray()[0];
this.featureAtModifyStart = feature.clone();
this.draggedCornerAtModifyStart = "";
feature.on("change", this.changeFeatureFunction);
}
private modifyEndFunction(event) {
const features = event.features;
const feature = features.getArray()[0];
feature.un("change", this.changeFeatureFunction);
// removing and adding feature to force reindexing
// of feature's snappable edges in OpenLayers
this.drawInteraction.features.clear();
this.drawInteraction.features.push(feature);
this.dispatchRettighetModifyEvent(feature);
}
changeFeatureFunction位于下方。只要用户仍在修改/拖动其中一个角,就会对几何所做的每个更改调用此函数。在这个函数中,我做了另一个函数,将修改后的矩形再次调整为矩形。这个“Rectanglify”功能移动与刚刚被用户移动的角落相邻的角落。
private changeFeatureFunction(event) {
let feature = event.target;
let geometry = feature.getGeometry();
// Removing change event temporarily to avoid infinite recursion
feature.un("change", this.changeFeatureFunction);
this.rectanglifyModifiedGeometry(geometry);
// Reenabling change event
feature.on("change", this.changeFeatureFunction);
}
没有太多细节,rectanglify功能需要
--
为了获得矩形的旋转,我们可以这样做:
export function getRadiansFromRectangle(feature: Feature): number {
const coords = getCoordinates(feature);
const point1 = coords[0];
const point2 = coords[1];
const deltaY = (point2[1] as number) - (point1[1] as number);
const deltaX = (point2[0] as number) - (point1[0] as number);
return Math.atan2(deltaY, deltaX);
}