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),用于围绕某个锚点旋转几何体。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该点:
我希望盒子/矩形能够保持其旋转,同时仍能够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?
答案 0 :(得分:0)
回答我想出的解决方案。
首先,将特征添加到ModifyInteraction中,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”函数看起来像这样。
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在下面。只要用户仍在修改/拖动角部之一,对几何体所做的每一次更改都将调用此函数。在此函数内部,我做了另一个函数,将修改后的矩形再次调整为矩形。此“重新排列”功能可移动与用户刚移动的角相邻的角。
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);
}
无需过多介绍,矩形函数就需要
-
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);
}