通过Javascript遍历复杂的JSON对象的最简单方法

Z时代
2024-01-10
分类：问答

我正在使用具有一些奇怪结构的JSON数据，例如：

{
    "RESULT": 
    {
        "COLUMNS": ["ID","name","ENABLED","perms","vcenabled","vcvalue","checkenabled","checkvalue","indxenabled","indxvalue"],
        "DATA": [
                    [7,"Site-A", 1, "1,2", 1, 1, 1, 0, 0, 0],
                    [15,"Site-B", 1, "1,2,3,4", 1, 1, 1, 0, 0, 0]
        ]
    },
    "ERROR": 0
}

我想创建一些JavaScript，将这些数据重组为适当的JSON结构，以使“列”数组值成为“数据”数组值的键。因此，在运行JS进程后，数据类似于以下内容：

[ {"ID":7,"name":"Site-A","ENABLED":1,"perms":"1,2","vcenabled":1,"vcvalue":1,"checkenabled":1,"checkvalue":1,"indxenabled":1,"indxvalue":1}, {"ID":15,"name":"Site-B","ENABLED":1,"perms":"1,2","vcenabled":1,"vcvalue":1,"checkenabled":1,"checkvalue":1,"indxenabled":1,"indxvalue":1} ]

完成JSON重组的JavaScript最佳做法是什么？我可以使用JQuery，Foundation JS等JS框架完成此任务吗？

回答：

newjson是您的新对象，j是您的json，

代码非常快，因为它缓存了很多信息，并且不使用push。

而且由于它是纯JavaScript，因此比所有库都快。

var j={
 "RESULT":{
  "COLUMNS":[
   "ID",
   "name",
   "ENABLED",
   "perms",
   "vcenabled",
   "vcvalue",
   "checkenabled",
   "checkvalue",
   "indxenabled",
   "indxvalue"
  ],
  "DATA":[
   [7,"Site-A", 1, "1,2", 1, 1, 1, 0, 0, 0],
   [15,"Site-B", 1, "1,2,3,4", 1, 1, 1, 0, 0, 0]
  ]
 },
 "ERROR": 0
}
var newjson=[],d=j.RESULT.COLUMNS.length;
for(var a=0,b=j.RESULT.DATA.length;a<b;a++){
 for(var c=0,tmpObj={};c<d;c++){
  tmpObj[j.RESULT.COLUMNS[c]]=j.RESULT.DATA[a][c];
 }
 newjson[a]=tmpObj;
}
console.log(newjson);

根据Bergi的响应，您还可以使用while–循环。

var orig={
 "RESULT":{
  "COLUMNS":[
   "ID",
   "name",
   "ENABLED",
   "perms",
   "vcenabled",
   "vcvalue",
   "checkenabled",
   "checkvalue",
   "indxenabled",
   "indxvalue"
  ],
  "DATA":[
   [7,"Site-A", 1, "1,2", 1, 1, 1, 0, 0, 0],
   [15,"Site-B", 1, "1,2,3,4", 1, 1, 1, 0, 0, 0]
  ]
 },
 "ERROR": 0
}
var formatted = [],
data = orig.RESULT.DATA,
cols = orig.RESULT.COLUMNS,
l = cols.length,
f = data.length;
while (f--) {
  var d = data[f],
      o = {},
      g = l;
  while (g--) {
    o[cols[g]] = d[g];
  }
  formatted[f] = o;
}