求高手 用js实现 用递归的方法得到如下的树形
var arr1 = {uId:"1",uName:"name1",uPass:"pass1",parentId:"0"};
arr.push(arr1);
var arr2 = {uId:"2",uName:"name2",uPass:"pass2",parentId:"1"};
arr.push(arr2);
var arr3 = {uId:"3",uName:"name3",uPass:"pass3",parentId:"1"};
arr.push(arr3);
var arr4 = {uId:"4",uName:"name4",uPass:"pass4",parentId:"2"};
arr.push(arr4);
var arr5 = {uId:"5",uName:"name5",uPass:"pass5",parentId:"3"};
arr.push(arr5);
var arr6 = {uId:"6",uName:"name6",uPass:"pass6",parentId:"3"};
arr.push(arr6);
var arr7 = {uId:"7",uName:"name7",uPass:"pass7",parentId:"4"};
arr.push(arr7);
用js实现
用递归的方法得到如下的树形
name1
|---name2
|---name4
|---name7
|---name3
|---name5
|---name6
追问:我要的不是这种多个for循环的代码,我需要的是一个递归的方法.
而且不是这种效果,我要的是一个树形结构.
答案:<body>
<div id="a"></div>
<script type="text/javascript">
var arr = new Array();
var arr1 = {uId:"1",uName:"name1",uPass:"pass1",parentId:"0"};
arr.push(arr1);
var arr2 = {uId:"2",uName:"name2",uPass:"pass2",parentId:"1"};
arr.push(arr2);
var arr3 = {uId:"3",uName:"name3",uPass:"pass3",parentId:"1"};
arr.push(arr3);
var arr4 = {uId:"4",uName:"name4",uPass:"pass4",parentId:"2"};
arr.push(arr4);
var arr5 = {uId:"5",uName:"name5",uPass:"pass5",parentId:"3"};
arr.push(arr5);
var arr6 = {uId:"6",uName:"name6",uPass:"pass6",parentId:"3"};
arr.push(arr6);
var arr7 = {uId:"7",uName:"name7",uPass:"pass7",parentId:"2"};
arr.push(arr7);
var str = "";
for(var i=0;i<arr.length;i++){
if(arr[i].parentId=="0"){
str = str+arr[i].uName+"<br/>";
for(var j=0;j<arr.length;j++){
if(arr[j].parentId==arr[i].uId){
str = str+" |--"+arr[j].uName+"<br/>";
for(var k=0;k<arr.length;k++){
if(arr[k].parentId==arr[j].uId){
str = str+" |--"+arr[k].uName+"<br/>";
}
}
}
}
}
}
document.getElementById("a").innerHTML = str;
</script>
</body>
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