通过Javascript对复杂JSON对象进行交互的最简单方法
作者:互联网
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,将这些数据重组为正确的JSON结构,以便“Column”数组值成为“DATA”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最佳实践是什么?我可以使用像JQuery,Foundation JS等JS框架来完成这项任务吗?
解决方法:
newjson是你的新对象,j是你的json,
代码非常快,因为它缓存了legth并且不使用push.
因为它是纯粹的javascript,它比所有库都要快.
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的反应你也可以使用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;
}
标签:javascript,jquery,json,javascript-framework 来源: https://codeday.me/bug/20191002/1844814.html