Question

我希望树中的所有单个节点都合并到其父节点中，但是如果其子节点包含两个或更多子节点，则不做任何更改。

这是我的数据模型示例：

[{
        name: "HOME",
        value: [{//Only node merged into the parent level
            name: "HOME",
            value: [{//Only node merged into the parent level
                name: "HOME",
                id: '1000'
            }] 
        }]
    }, {
        name: "ARTICLE",
        value: [{
            name: "ARTICLE",
            value: [{
                name: "ARTICLE TYPE 1",
                id: '2001'
            },{
                name: "ARTICLE TYPE 2",
                id: '2002'
            }] 
        },{
            name: "ARTICLE",
            value: [{//Only node merged into the parent level
                name: "ARTICLE TYPE 3",
                id: '2003'
            }] 
        }]
    }]

我想像这样过滤数据：

    [{
        name: "HOME",
        id: 1000
    }, {
        name: "ARTICLE",
        value: [{
            name: "ARTICLE",
            value: [{
                name: "ARTICLE TYPE 1",
                id: '2001'
            },{
                name: "ARTICLE TYPE 2",
                id: '2002'
            }] 
        },{
            name: "ARTICLE TYPE 3",
            id: '2003'
        }]
    }]

//更新1：这是个想法，但是现在有一个问题，发现节点无法回退到原始节点，只能修改当前的父节点：

function filter(data){
    for(var i = 0; i < data.length; i++){
        if( !data[i].value ) continue;

        //Check whether there are child nodes "value" is because it contains a "value" does not exist "id",
        //you must enter a recursive make the following checks
        if( data[i].value.length === 1 && !data[i].value[0].value ) {
            data[i].id = data[i].value[0].id;
            delete data[i].value;
            continue;
        }
        filter( data[i].value );
    }
    return data;
}

我现在直接修改原始对象，我不知道这样做是否合理。

//更新2：我的最终措辞是这样的，结果输出看起来是正确的，但是不确定逻辑是否正确，看起来非常难看，或者不知道是否有更好的解决方案？

function filter(data, parent){
    for(var i = 0; i < data.length; i++){
        if( data[i].value ) filter( data[i].value, data[i] );

        if( parent && data.length === 1 && !data[i].value ) {
            parent.id = data[i].id;
            delete parent.value;
        }
    }
    return data;
}

Answer 1

一个简单的版本

function merge(node){
    if(node.value){
        var children = node.value.map(merge);
        return children.length === 1?
            children[0]:
            {
                name: node.name,
                value: children
            };
    }
    return node;
}
var result = data.map(merge);

左右：

function cp(a, b){
    for(var k in b){
        if(k === "value" || k in a) continue;
        a[k] = b[k];
    }
    return a;
}

function merge(node){
    if(node.value){
        var children = node.value.map(merge);
        return children.length===1 && !("value" in children[0])?
            cp(children[0], node):
            cp({ value: children }, node);
    }
    return cp({}, node);
}
var result = data.map(merge);

Answer 2

这个版本更容易理解。功能要么

没有值属性时不执行任何操作
只有一个孩子时与孩子合并
在有多个孩子的情况下为每个孩子重复

＆＃13;

var data = [{
        name: "HOME",
        value: [{//Only node merged into the parent level
            name: "HOME",
            value: [{//Only node merged into the parent level
                name: "HOME",
                id: '1000'
            }] 
        }]
    }, {
        name: "ARTICLE",
        value: [{
            name: "ARTICLE",
            value: [{
                name: "ARTICLE TYPE 1",
                id: '2001'
            },{
                name: "ARTICLE TYPE 2",
                id: '2002'
            }] 
        },{
            name: "ARTICLE",
            value: [{//Only node merged into the parent level
                name: "ARTICLE TYPE 3",
                id: '2003'
            }] 
        }]
    }];
    
function merge(node) {
  if (node.value == undefined) return;
  if (node.value.length == 1) {
    var child = node.value[0];
    node.name = child.name;
    if (child.id) node.id = child.id;
    if (child.value) node.value = child.value;
    else delete node.value;
    merge(node);
  } else {
    node.value.forEach(function(child) {
      merge(child);
    });
  }
}

$('#input').html(JSON.stringify(data));
data.forEach(function(node) { merge(node) });
$('#output').html(JSON.stringify(data));

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
Input: <div id="input"></div>
Output: <div id="output"></div>

＆＃13;

如何使用递归过滤所有单个JSON数据？

2 个答案: