Javascript 根据多个关键点合并两个数组中的对象

下面我有两个数组，我想基于两个常用键（“iso3”和“year”）合并它们。为了合并，每个对象必须同时具有这两个对象。数组的长度不同

array1 = [{
  "id":24006,
  "iso3":"AFG",
  "country":"Afghanistan",
  "year":2014,
  "value":29.78
},
{
  "id":138806,
  "iso3":"ALB",
  "country":"Albania",
  "year":2013,
  "value":0.6341109715
},
{
  "id":44206,
  "iso3":"DZA",
  "country":"Algeria",
  "year":2014,
  "value":39.928947
}]

array2 = [{
  "indicator_id":21806,
  "footnote_id":64811,
  "iso3":"AFG",
  "year":2014
},
{
  "indicator_id":23806,
  "footnote_id":15711,
  "iso3":"AFG",
  "year":2013
},
{
  "indicator_id":123406,
  "footnote_id":15711,
  "iso3":"ALB",
  "year":2013
},
{
  "indicator_id":101606,
  "footnote_id":48911,
  "iso3":"DZA",
  "year":2013
}];

---

，但数组仅基于一个公共密钥合并。我试着用那个代码工作，但没有成功。感谢您的帮助

您需要做的是扩展您所在站点的答案中提供的加入条件。像这样：

var combined = [];
function findSecond(iso3, year, second){
    for (var i = 0; i < second.length; i += 1){
    //Adding the second condition
        if (second[i].iso3 === iso3 && second[i].year ===  year ) {
            return second[i];
        }
    }
    return null;
}

while (el = array1.pop()){
    var getSec = findSecond(el.iso3, el.year, array2);
    if (getSec){
        for (var l in getSec){
            if (!(l in el)) {
                el[l] = getSec[l];
            }
        }
        combined.push(el);
    }
}

var组合=[]；
功能查找秒（iso3，年，秒）{
对于（变量i=0；i

另一种利用和

代码：

---

这种方法首先基于键创建合并对象的字典（或映射）。它使用jQuery$.extend合并单个对象（其他实用程序库可能会提供合并对象的方法）

生成字典后，它会转换回数组，并通过单个对象键进行过滤，以生成类似内部联接的结果。如果要进行外部联接，只需删除筛选器

var array1 = [{
  "id":24006,
  "iso3":"AFG",
  "country":"Afghanistan",
  "year":2014,
  "value":29.78
},
{
  "id":138806,
  "iso3":"ALB",
  "country":"Albania",
  "year":2013,
  "value":0.6341109715
},
{
  "id":44206,
  "iso3":"DZA",
  "country":"Algeria",
  "year":2014,
  "value":39.928947
}]

var array2 = [{
  "indicator_id":21806,
  "footnote_id":64811,
  "iso3":"AFG",
  "year":2014
},
{
  "indicator_id":23806,
  "footnote_id":15711,
  "iso3":"AFG",
  "year":2013
},
{
  "indicator_id":123406,
  "footnote_id":15711,
  "iso3":"ALB",
  "year":2013
},
{
  "indicator_id":101606,
  "footnote_id":48911,
  "iso3":"DZA",
  "year":2013
}];

// merge by key where possible
var dict = array1.concat(array2).reduce(function (dict, val) {
    // Use the two values we want to join on as a key
      var key = val.iso3 + val.year;

    // Lookup existing object with same key, default to
    // empty object if one does not exist
    var existing = dict[key] || {};

    // Merge two objects
    dict[key] = $.extend(existing, val);

    return dict; 
}, {});

var joined = Object.keys(dict).map(function (key) {
  // pull the dictionary values back out into an array
  return dict[key];
}).filter(function (obj) {
  // this is making it work like an inner out, an outer join can be
  // done by removing the filter
  return (typeof obj.indicator_id !== "undefined") && (typeof obj.country !== "undefined");
});

---

JS Fiddle（输出记录到控制台：）

此解决方案具有一个用于收集项目的对象，然后呈现所需的数组

var array1=[{“id”：24006，“iso3”：“AFG”，“国家”：“阿富汗”，“年份”：2014，“价值”：29.78}，{“id”：138806，“iso3”：“ALB”，“国家”：“阿尔巴尼亚”，“年份”：2013，“价值”：0.6341109715}，{“id”：44206，“iso3”：“DZA”，“国家”：“阿尔及利亚”，“年份”：2014，“价值”：39.928947}，
array2=[{“指标id”：21806，“脚注id”：64811，“iso3”：“AFG”，“年份”：2014}，{“指标id”：23806，“脚注id”：15711，“iso3”：“AFG”，“年份”：2013}，{“指标id”：123406，“脚注id”：15711，“iso3”：“ALB”，“年份”：2013}，{“指标id”：101606，“脚注id”：48911，“iso3”：“DZA”，“年份”：2013}；
函数合并（array1，array2）{
函数makeObj（a）{
obj[a.iso3]=obj[a.iso3]| |{}；
obj[a.iso3][a.year]=obj[a.iso3][a.year]|{}；
Object.keys（a）.forEach（函数（k）{
obj[a.iso3][a.year][k]=a[k]；
});
}
变量数组=[]，
obj={}；
阵列1.forEach（makeObj）；
array2.forEach（makeObj）；
Object.keys（obj.forEach）（函数（k）{
Object.key（obj[k]）.forEach（function（kk）{
array.push（obj[k][kk]）；
});
});
返回数组；
}
write（“”+JSON.stringify（merge（array1，array2），0，4）+’

因此您需要从具有相同年份和ISO的数组中筛选出对象。是这样吗？根据您链接到的答案中的示例，编辑行

if（second[i].id==id）{

to

if（second[i].iso3==iso3&second[i].year==year）{

试试看。看看是否有用？@HarryBomrah是的，没错。@Irtza.QC这是我已经尝试过的，它给了我一个空白屏幕。我做了一个JSFIDLE，但出现了错误。我不确定我遗漏了什么。你能看看吗？你的小提琴工作得很好：它连接了两个记录。在这里，在结尾添加了一个简单的测试：这很完美。谢谢你太感谢你了！哇，谢谢你把我推过了15k积分的边缘；）

[{
  "indicator_id": 23806,      /* Found in array2 only, not merged */
  "footnote_id": 15711,
  "iso3": "AFG",
  "year": 2013
}, {
  "id": 138806,               /* This one has been merged */
  "iso3": "ALB",
  "country": "Albania",
  "year": 2013,
  "value": 0.6341109715,
  "indicator_id": 123406,
  "footnote_id": 15711
}, {           
  "indicator_id": 101606,     /* Found in array2 only, not merged */
  "footnote_id": 48911,
  "iso3": "DZA",
  "year": 2013
}, {
  "id": 24006,                /* This one has been merged */
  "iso3": "AFG",
  "country": "Afghanistan",
  "year": 2014,
  "value": 29.78,
  "indicator_id": 21806,
  "footnote_id": 64811
}, {
  "id": 44206,                /* Found in array1 only, not merged */
  "iso3": "DZA",
  "country": "Algeria",
  "year": 2014,
  "value": 39.928947
}]

var array1 = [{
  "id":24006,
  "iso3":"AFG",
  "country":"Afghanistan",
  "year":2014,
  "value":29.78
},
{
  "id":138806,
  "iso3":"ALB",
  "country":"Albania",
  "year":2013,
  "value":0.6341109715
},
{
  "id":44206,
  "iso3":"DZA",
  "country":"Algeria",
  "year":2014,
  "value":39.928947
}]

var array2 = [{
  "indicator_id":21806,
  "footnote_id":64811,
  "iso3":"AFG",
  "year":2014
},
{
  "indicator_id":23806,
  "footnote_id":15711,
  "iso3":"AFG",
  "year":2013
},
{
  "indicator_id":123406,
  "footnote_id":15711,
  "iso3":"ALB",
  "year":2013
},
{
  "indicator_id":101606,
  "footnote_id":48911,
  "iso3":"DZA",
  "year":2013
}];

// merge by key where possible
var dict = array1.concat(array2).reduce(function (dict, val) {
    // Use the two values we want to join on as a key
      var key = val.iso3 + val.year;

    // Lookup existing object with same key, default to
    // empty object if one does not exist
    var existing = dict[key] || {};

    // Merge two objects
    dict[key] = $.extend(existing, val);

    return dict; 
}, {});

var joined = Object.keys(dict).map(function (key) {
  // pull the dictionary values back out into an array
  return dict[key];
}).filter(function (obj) {
  // this is making it work like an inner out, an outer join can be
  // done by removing the filter
  return (typeof obj.indicator_id !== "undefined") && (typeof obj.country !== "undefined");
});