JsPlumb端点叠加

Question

假设我有一个有4个端点的元素，每个端点都有一个连接。我需要显示端点标签中的所有连接，并通过单击其名称删除任何连接。jsPlumb开箱即有这个能力吗？或者我该怎么做呢？

​

sourceEndpoint = jsPlumb.addEndpoint($(requirementSelector), {overlays: removeLabel, maxConnections: -1, endpoint: ["Dot", { radius: 4}], anchors: ["RightMiddle", "LeftMiddle"]});
    targetEndpoint = jsPlumb.addEndpoint($(solutionSelector), {overlays: removeLabel,maxConnections: -1, endpoint: ["Dot", { radius: 4}], anchors: ["RightMiddle", "LeftMiddle"]});

    jsPlumb.connect({
        source: sourceEndpoint, 
        target: targetEndpoint 
    });

    targetEndpoint.bind("click", function(endpoint) {
      var elementEndpoints = jsPlumb.selectEndpoints({element: endpoint.elementId});

      var ids="<div style='border: 2px solid black; padding: 5px; background-color: #ffffff'; z-index:10;>";

      elementEndpoints.each(function(ep){
        ids += "<p ng-click='clicked()'>Remove - " + ep.id + "</p>"
      });

      ids += "</div>";

      endpoint.setLabel(ids);
      endpoint.showOverlay();
    });

Answer 1

尝尝这个。每次创建新连接时，绑定一个事件以删除该连接：

jsPlumb.bind("jsPlumbConnection", function(ci) {
            ci.connection.bind("click",function(con){
                jsPlumb.detach(con);
            });
        });

让我知道它是否适用于您。