Question

我有四个数组：

var attributes = ["a","b","c","d"];
var a = [1,2,3];
var b = [34,55,66];
var c = [22,23,53];
var d = [15,78,98];

如何在Nodejs中将它们合并为一个类似字符串的json？

[{"a":1,"b":34,"c":22,"d":14},
 {"a":2,"b":55,"c":23,"d":78},
 {"a":3,"b":66,"c":53,"d":98}]

这是我的代码，但是有人有更好的解决方案吗？我确实需要保留报价。

＆＃13;

var a = [1,2,3];
var b = [34,55,66];
var c = [22,23,53];
var d = [15,78,98];

var obj = "[";
for (var u = 0; u < a.length; u++) {
  var l = "\"a\":"+a[u]+",";
  var m = "\"b\":"+b[u]+",";
  var q = "\"b\":"+c[u]+",";
  var n = "\"d\":"+d[u]+"";
  if(u == (a.length-1))
    var k = "{" + l + m + q + n + "}";
  else
    var k = "{" + l + m + q + n + "},";
  console.log(k);
  obj = obj + k;
};
obj = obj + "]";
console.log(obj);

＆＃13;

Answer 1

假设所有数组的长度相同，并对其名称进行硬编码：

   var obj = []
    for (var u = 0; u < a.length; u++) {
        obj.push({
            'a': a[u],
            'b': b[u],
            'c': c[u],
            'd': d[u]
        });
    };
    obj = JSON.stringify(obj);

编辑：将obj转换为json字符串，该问题被错误地编辑以询问数组。

Answer 2

这段代码可以解决问题：

var arrays = [a, b, c, d]; // add all the arrays you want
var num = a.length; // or hardcode to the length you want

var result = [];
for(var i = 0; i < num; i++) {
    var element = {};
    attributes.forEach(function(attr, index) {
        element[attr] = arrays[index][i];
    });
    result.push(element);
}

＆＃13;

var attributes = ["a","b","c","d"];
var a = [1,2,3];
var b = [34,55,66];
var c = [22,23,53];
var d = [15,78,98];

var arrays = [a, b, c, d];
var result = [];
var num = a.length;
for(var i = 0; i < num; i++) {
    var element = {};
    attributes.forEach(function(attr, index) {
        element[attr] = arrays[index][i];
    });
    result.push(element);
}

document.write(JSON.stringify(result));

＆＃13;

Answer 3

var attributes = ["a","b","c","d"];
var a = [1,2,3];
var b = [34,55,66];
var c = [22,23,53];
var d = [15,78,98];

var x = [];

for (var i=0;i<3;i++) {
  var obj = {};
  for (var j=0;j<attributes.length;j++){
    obj[attributes[j]]=eval(attributes[j]+"["+i+"]");
  }
  x.push(obj);
}
console.log(x);

Answer 4

我知道这个问题已有多个答案，但似乎没有人处理关键数组，因为变量的名称是在属性数组中引入的。由于我认为这是属性数组的要点，我的解决方案就是使用节点的全局范围。它也不假设所有的键数组都与未指定的相同（即使它们是示例）。

var attributes = ["a","b","c","d"];
var a = [1,2,3];
var b = [34,55,66];
var c = [22,23,53];
var d = [15,78,98];
var arr = [];
while(true){
    var obj = {};
    for(var i=0;i<attributes.length;i++){
    if(global[attributes[i]].length)obj[attributes[i]]=global[attributes[i]].shift();
  }
  if(Object.keys(obj).length)
    arr.push(obj);
  else
    break;
}

在下面的小提琴中，我不得不通过手动创建全局数组来模拟节点的全局范围，但是你明白了这一点：

小提琴：https://jsfiddle.net/trex005/vm1yeL5d/

重要提示如果数组实际上不在全局范围内，但在某些其他对象中，则可以使用该对象代替global

将数组转换为子节点，如Nodejs中的对象字符串

4 个答案: