JSON對象-遞歸更改所有子代的屬性名稱

Question

我有以下json對象。

myJsonObj = {  
   "name":"Laptop",
   "type":"hardware computer laptop",
   "forward":[  
      {  
         "name":"Depends On",
         "forward":[  
            {  
               "name":"test asset 1",
               "type":"hardware",
               "link":"somelink"
            }
         ]
      },
      {  
         "name":"somename",
         "forward":[  
            {  
               "name":"test asset 5",
               "type":"hardware",
               "link":"somelink"
            }
         ]
      }
   ],
   "inverse":[  
      {  
         "name":"somename",
         "inverse":[  
            {  
               "name":"test asset 4",
               "ciTypeCls":"hardware",
               "link":"somelink"
            },
            {  
               "name":"test asset 1",
               "ciTypeCls":"hardware",
               "link":"somelink"
            }
         ]
      },
      {  
         "name":"somename",
         "inverse":[  
            {  
               "name":"test asset 1",
               "ciTypeCls":"hardware",
               "link":"somelink"
            }
         ]
      }
   ]
}

我試圖改變屬性名稱forward到children 。

以下是我的代碼

 myJsonObj = { "name": "Laptop", "type": "hardware computer laptop", "forward": [{ "name": "Depends On", "forward": [{ "name": "test asset 1", "type": "hardware", "link": "somelink" }] }, { "name": "somename", "forward": [{ "name": "test asset 5", "type": "hardware", "link": "somelink" }] }], "inverse": [{ "name": "somename", "inverse": [{ "name": "test asset 4", "ciTypeCls": "hardware", "link": "somelink" }, { "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }, { "name": "somename", "inverse": [{ "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }] } jQuery.each(myJsonObj, function(e) { e.children = e.forward; delete e.forward; }); console.log(myJsonObj); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

它似乎不起作用，我在做什么錯？

Answer 1

嘗試這個：

 jQuery.each(myJsonObj.forward, function(i , item) {
     //Nested forward
     item.children = item.forward;
     delete item.forward;
});
   //PARENT forward
    myJsonObj.children = myJsonObj.forward;
    delete myJsonObj.forward;

JSFiddle https://jsfiddle.net/xbkp10Ly/

Answer 2

根據您當前的JSON結構，簡單的“查找和替換”即可工作。

 var myJsonObj = { "name": "Laptop", "type": "hardware computer laptop", "forward": [{ "name": "Depends On", "forward": [{ "name": "test asset 1", "type": "hardware", "link": "somelink" }] }, { "name": "somename", "forward": [{ "name": "test asset 5", "type": "hardware", "link": "somelink" }] }], "inverse": [{ "name": "somename", "inverse": [{ "name": "test asset 4", "ciTypeCls": "hardware", "link": "somelink" }, { "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }, { "name": "somename", "inverse": [{ "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }] } document.write('<pre>' + JSON.stringify(myJsonObj).replace(/forward/g, "children") + '</pre>'); 

更安全，因為它只會檢查屬性而不是價值

JSON.parse(JSON.stringify(myJsonObj).replace(/forward(?=\":)/g,"children"))

檢查此鏈接

Answer 3

這是一個以遞歸樣式對對象和數組進行迭代的提議。

它使用深度優先搜索 ，因為實際節點必須保留，直到訪問所有從屬節點為止。

 var obj = { "name": "Laptop", "type": "hardware computer laptop", "forward": [{ "name": "Depends On", "forward": [{ "name": "test asset 1", "type": "hardware", "link": "somelink" }] }, { "name": "somename", "forward": [{ "name": "test asset 5", "type": "hardware", "link": "somelink" }] }], "inverse": [{ "name": "somename", "inverse": [{ "name": "test asset 4", "ciTypeCls": "hardware", "link": "somelink" }, { "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }, { "name": "somename", "inverse": [{ "name": "test asset 1", "ciTypeCls": "hardware", "link": "somelink" }] }] }; function replaceProperty(o, from, to) { function f(r, p) { if (Array.isArray(r[p])) { iterA(r[p]); } if (typeof r[p] === 'object') { iterO(r[p]); } if (p === from) { r[to] = r[from]; delete r[from]; } } function iterO(o) { Object.keys(o).forEach(function (k) { f(o, k); }); } function iterA(a) { a.forEach(function (_, i, aa) { f(aa, i); }); } iterO(o); } replaceProperty(obj, 'forward', 'children'); document.write('<pre>' + JSON.stringify(obj, 0, 4) + '</pre>');