代码之家 › 专栏 › 技术社区 › bob.mazzo

递归的javascript函数,通过其uid定位对象

recursion javascript

bob.mazzo · 技术社区 · 7 年前

例如,如果我选择的项目是:

{
 "UID": 49,
 "GUID": "",
 "LocationName": "Doctor Smith's Office",

 "LocationType": {
    "UID": 2,
    "LocationTypeName": "Practice",
    "Description": "other location"
 }
}

UID

{
  UID: 2,
  GUID: "",
  LocationName: "USA",
  ParentLocation: null,
  subs: [{
	UID: 42,
	GUID: "",
	LocationName: "New Jersey",
	Description: "",
	subs: [{
		UID: 3,
		GUID: "",
		LocationName: "Essex County",
		ParentLocation: null,
		"subs":[
			UID: 4,
			LocationName: "Newark",
			ParentLocation: 3,
			"subs": [
				{
					"UID": 49,
					"GUID": "",
					"LocationName": "Doctor Smith's Office",
													
					"LocationType": {
						"UID": 2,
						"LocationTypeName": "Practice",
						"Description": "other location"
					},                                    
					"subs": [
						{
							"HostID": 38,
							"HostName": "Ocean Host",
						}
					]
				}
			]
		]
	  }                  
	]
  }]
};

let foundItem = this.findInTreeView(this.treeviewData[0], node.selectedNode);

// find selected node in treeview nav
	// param: data - the treeview dataset
	// param: selected - the selected node to be searched for in param 'data'
	findInTreeView(data: any, selected: any )	{		
		let found;
		if (this.foundInTree(data, selected)) {			
			return data;
		}
		let elem;
		let ary = data.subs;
		for (var i=0; i < ary.length; i++) {
			elem = ary[i];
			if (this.foundInTree(elem, selected)) {		
                // *** PROBLEM: If func has return true, I want to return the 'elem' object.	
				return elem;
			}
		}
		for (var i=0; i < ary.length; i++) {
			elem = ary[i];
			if (elem.subs !== undefined) { 
				// recurse subs array
				let found = this.findInTreeView(elem, selected); 
				if (found) {
					return elem;
				}
			}
		}	
		//return elem;		
	}
  
  foundInTree(treeItem, node) {
		if (treeItem.UID === node.UID) {
			return true;			
		}
		else {
			return false;
		}
	}

3 回复 | 直到 7 年前

CertainPerformance 7 年前

使用递归更容易 reduce

const input={UID:2,GUID:"",LocationName:"USA",ParentLocation:null,subs:[{UID:42,GUID:"",LocationName:"New Jersey",Description:"",subs:[{UID:3,GUID:"",LocationName:"Essex County",ParentLocation:null,"subs":[{UID:4,LocationName:"Newark",ParentLocation:3,"subs":[{"UID":49,"GUID":"","LocationName":"Doctor Smith's Office","LocationType":{"UID":2,"LocationTypeName":"Practice","Description":"other location"},"subs":[{"HostID":38,"HostName":"Ocean Host",}]}]}]}]}]};

const findUIDObj = (uid, parent) => {
  const { UID, subs } = parent;
  if (UID === uid) {
    const { subs, ...rest } = parent;
    return rest;
  }
  if (subs) return subs.reduce((found, child) => found || findUIDObj(uid, child), null);
};
console.log(findUIDObj(49, input))

Nina Scholz 7 年前

您可以使用一个显式函数来搜索所需的 UID

function find(array, UID) {
    var object;
    array.some(o => {
        if (o.UID === UID) {
            return object = o;
        }
        return object = find(o.subs, UID);
    });
    return object;
}

var object = { UID: 2, GUID: "", LocationName: "USA", ParentLocation: null, subs: [{ UID: 42, GUID: "", LocationName: "New Jersey", Description: "", subs: [{ UID: 3, GUID: "", LocationName: "Essex County", ParentLocation: null, subs: [{ UID: 4, LocationName: "Newark", ParentLocation: 3, subs: [{ UID: 49, GUID: "", LocationName: "Doctor Smith's Office", LocationType: { UID: 2, LocationTypeName: "Practice", Description: "other location" }, subs: [{ HostID: 38, HostName: "Ocean Host", }] }] }] }] }] };

console.log(find([object], 49));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Scott Sauyet 7 年前

实现这一点的一种方法是编写一个相当通用的树查找函数版本,然后针对特定的问题对其进行配置。在这里,我们选择通过匹配提供的uid进行测试,我们通过查看 subs 潜艇

const searchTreeDF = (kids, test, convert, node) => test(node) // depth-first search
  ? convert(node) 
  : (kids(node) || []).reduce(
      (found, child) => found || searchTreeDF(kids, test, convert, child), 
      false
  )

const subs = node => node.subs
const matchId = (uid) => (item) => item.UID === uid
const convert = ({subs, ...rest}) => ({...rest})

const findUid = (uid, tree) => searchTreeDF(subs, matchId(uid), convert, tree)
// ...
const tree = {"GUID": "", "LocationName": "USA", "ParentLocation": null, "UID": 2, "subs": [{"Description": "", "GUID": "", "LocationName": "New Jersey", "UID": 42, "subs": [{"GUID": "", "LocationName": "Essex County", "ParentLocation": null, "UID": 3, "subs": [{"LocationName": "Newark", "ParentLocation": 3, "UID": 4, "subs": [{"GUID": "", "LocationName": "Doctor Smith's Office", "LocationType": {"Description": "other location", "LocationTypeName": "Practice", "UID": 2}, "UID": 49, "subs": [{"HostID": 38, "HostName": "Ocean Host"}]}]}]}]}]}


console.log(findUid(49, tree))

UID 直接传递,但要传递的元素属性,我们可以写

const matchElem = (elem) => (item) => elem.UID === item.UID

const findUid2 = (elem, tree) => searchTreeDF(subs, matchElem(elem), convert, tree)
// ...
findUid2({UID: 49}, tree)

或者如果我们不想转换结果,并保留 convert 以下内容:

const findUid = (uid, tree) => searchTreeDF(subs, matchId(uid), x => x, tree)

const findUid = (uid, tree) => searchTreeDF(
  node => node.subs || [], 
  (item) => item.UID === uid,
  ({subs, ...rest}) => ({...rest}), 
  tree
)