从扁平结构中获取JSON的每个名称值

时间:2016-02-26 07:50:17

标签: javascript json structure

我有一个JSON文件,其数据结构如下:

{
    "first": {
        "second": "example",
        "third": {
            "fourth": "example2",
            "fifth": "example3"
        }
    }
}

有没有办法将它转换为扁平结构,只能获得带字符串值的名称 - 值对?从这个JSON我想得到这样的东西:

{
"second": "example",
"fourth": "example2",
"fifth": "example3"
}

2 个答案:

答案 0 :(得分:2)

这可能会让您在纯JavaScript中了解如何展平对象。这很粗糙,但可以扩展:

function flatten(obj) {
    var flattened = {};

    for (var prop in obj)
        if (obj.hasOwnProperty(prop)) {
            //If it's an object, and not an array, then enter recursively (reduction case).
            if (typeof obj[prop] === 'object' && 
                Object.prototype.toString.call(obj[prop]) !== '[object Array]') {
                var child = flatten(obj[prop]);

                for (var p in child)
                   if (child.hasOwnProperty(p))
                       flattened[p] = child[p];
            }
            //Otherwise if it's a string, add to our flattened object (base case).
            else if (typeof obj[prop] === 'string')
                flattened[prop] = obj[prop];
        }

    return flattened;
}

fiddle

答案 1 :(得分:2)

可以通过递归函数完成:

var obj = {
"first": {
    "second": "example",
    "third": {
        "fourth": "example2",
        "fifth": "example3"
    }
}
};


function parseObj(_object) {
    var tmp = {};
    $.each(_object, function(k, v) {
  if(typeof v == 'object') {
    tmp = $.extend({}, tmp, parseObj(v));
  } else {
    tmp[k] = v; 
  }

});
    return tmp;
}

var objParsed = {};
objParsed = parseObj(obj);
console.log(objParsed);

这是JSFiddle的工作:https://jsfiddle.net/0ohbyu7b/