如何使用遞歸篩選所有單個JSON數據？

Question

我希望所有的樹都合併到其父的各個節點，但如果它的子節點包含兩個或兩個以上的子節點，沒有任何變化

這裏是我的數據模型，例如：

[{ 
     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' 
      }] 
     }] 
    }] 

我要過濾的數據得到這樣的：

[{ 
     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： 這是想法，但是現在有一個問題，找到節點不能回下降到原來的節點，您只能修改cu rrent父節點：

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; 
} 

我現在直接修改原始對象，我不知道這樣做是否合理。

//更新2： 我的最終措辭是這樣的，結果輸出看起來是正確的，但不確定邏輯是否正確，並且看起來非常難看，或者不知道是否有更好的解決方案？

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

簡單版

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

這個版本更容易理解。該函數或者

• 一籌莫展的時候，它沒有價值屬性
• 合併有孩子時，它只有一個孩子
• 重複的每一個孩子，當有多個孩子

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));

<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>