Javascript 从平均堆栈应用程序的控制器中的范围变量中选择不同的值

如何从$scope.items单独获取不同的建筑ID列表并存储到$scope.destinationBuildingIdList变量中？

只需使用javascript映射函数即可

$scope.items = [
{
sourceBuildingId: "B001",
sourceBuildingName: "Building 1",
destinationBuildingId: "B002"
destinationBuildingName: "Building 2" 
},
{
sourceBuildingId: "B003",
sourceBuildingName: "Building 3",
destinationBuildingId: "B002",
destinationBuildingName: "Building 2" 
},
{
sourceBuildingId: "B001",
sourceBuildingName: "Building 1",
destinationBuildingId: "B003",
destinationBuildingName: "Building 3" 
},
{
sourceBuildingId: "B002",
sourceBuildingName: "Building 2",
destinationBuildingId: "B004",
destinationBuildingName: "Building 4" 
}];

$scope.destinationBuildingIdList="";

演示

角度。模数PP，[] .controllerctrl，函数$scope{ $scope.items=[ { sourceBuildingId:B001， sourceBuildingName:1号楼， destinationBuildingId:B002， 目的地建筑物名称：2号建筑物 }, { sourceBuildingId:B003， sourceBuildingName:3号楼， destinationBuildingId:B002， 目的地建筑物名称：2号建筑物 }, { sourceBuildingId:B001， sourceBuildingName:1号楼， destinationBuildingId:B003， 目的地建筑物名称：3号建筑物 }, { sourceBuildingId:B002， sourceBuildingName:2号楼， destinationBuildingId:B004， 目的地建筑物名称：4号建筑物 }]; $scope.destinationBuildingIdList=$scope.items.mapo=>o.destinationBuildingId console.log$scope.destinationBuildingIdList }

---

Array.fromnew Set$scope.items.mapo=>o.destinationBuildingId以获得唯一值。@AFP_555太好了，效果很好。@Sajeetharan所选的重复问题只解决数组中不同的值。它没有解释如何根据其他列出的对象的属性创建列表，正如Sachila的回答所示。关闭此链接对吗？@AFP_555问题的主要部分与我所附的链接类似。休息是分配的问题。因此它是重复的。@Sajeetharan只是一个问题，因为我真的不知道，一个问题可以分配多个重复的吗？

$scope.destinationBuildingIdList= $scope.items.map(o=> o.destinationBuildingId)