Question

我正在尝试创建一个从两个独立对象获取信息的对象（取自ajax调用）。基本上我们有一个标签列表，还有一个元素列表，这些项是相关的，但是标签对象不包含我填充所有数据所需的所有信息。例如：

标签列表：

[
    {"id": 1, "name": "yadda", "description": "yadda yadda"},
    {"id": 2, "name": "yadda1", "description": "yadda yadda1"},
    {"id": 7, "name": "yadda2", "description": "yadda yadda2"},
    {"id": 10, "name": "yadda3", "description": "yadda yadda3"}
]

元素列表（更多信息）：

[
    {"id": 1, "icon": "icon1.gif"},
    {"id": 2, "icon": "icon2.gif"},
    {"id": 7, "icon": "icon3.gif"},
    {"id": 10, "icon": "icon4.gif"}
]

我需要通过ID比较两个对象，并将它们组合成一个新对象，使我能够访问这两个对象中的所有数据。

如果它有帮助，这是一个有角度的项目，我已经在使用下划线了，我相信必须有一些方法用下划线来做到这一点，但我不是很熟悉它。

Answer 1

又快又脏：

var a = [
        {"id": 1, "name": "yadda", "description": "yadda yadda"},
        {"id": 2, "name": "yadda1", "description": "yadda yadda1"},
        {"id": 7, "name": "yadda2", "description": "yadda yadda2"},
        {"id": 10, "name": "yadda3", "description": "yadda yadda3"}
    ], 
    b = [
        {"id": 1, "icon": "icon1.gif"},
        {"id": 2, "icon": "icon2.gif"},
        {"id": 7, "icon": "icon3.gif"},
        {"id": 10, "icon": "icon4.gif"}
    ];

var result = a.map(function(v){

    var ret;

    $.each(b, function(k, v2){

        if(v2.id === v.id){
            ret = $.extend({}, v2, v); // merge the objects in to a new one
            return false; // break the loop
        }      

    });

    return ret;

});

console.log(result);

http://jsfiddle.net/YwUA2/

如您所见，这假设两个数组中的对象之间存在1：1的关系。

Answer 2

对于它的价值，没有图书馆，只有香草js

完成工作的功能：

function mergeObjArrays(list1, list2) {
    var mergedList = [],
    i = 0,
    j = 0,
    k = 0,
    l1 = list1.length,
    l2 = list2.length,
    listMatchFound = false,
    mergedObj,
    mergeMatchFound = false;
    for (i = 0; i < l1; i++) {
        for (j = 0; j < l2; j++) {
            if (list1[i].id === list2[j].id) {
                listMatchFound = true;
                mergedObj = mergeObj(list1[i], list2[j]);
                for (k = 0; k < mergedList.length; k++) {
                    if (mergedList[k].id === mergedObj.id) {
                        mergedObj = mergeObj(mergedList[k], mergedObj);
                        mergedList[k] = mergedObj;
                        mergeMatchFound = true;
                        break;
                    }
                }
                if (!mergeMatchFound) {
                    mergedList.push(mergedObj);
                } else {
                    mergeMatchFound = false; //reset ready for another iteration
                }
            }
        }
        if (!listMatchFound) {
            mergedList.push(list1[i]);
        } else {
            listMatchFound = false; //reset ready for another iteration
        }
    }
    return mergedList;

    function mergeObj(obj1, obj2) {
        for (var o in obj1) {
            obj2[o] = obj1[o];
        }
        return obj2;
    }
}

使用示例

var a = [
        {"id": 1, "name": "yadda", "description": "yadda yadda"},
        {"id": 2, "name": "yadda1", "description": "yadda yadda1"},
        {"id": 7, "name": "yadda2", "description": "yadda yadda2"},
        {"id": 10, "name": "yadda3", "description": "yadda yadda3"}
    ];

var b = [    
        {"id": 7, "icon": "icon3.gif"},
        {"id": 10, "icon": "icon4.gif"},
        {"id": 2, "icon": "icon2.gif"},
        {"id": 2, "title": "Number 2 title"},
        {"id": 1, "icon": "icon1.gif"}
    ];

//used like

var result = mergeObjArrays(a,b);

console.log(result)

//outputs (although not necessarily in order)
[
 {"id": 1, "name": "yadda", "description": "yadda yadda","icon": "icon1.gif"},
 {"id": 2, "name": "yadda1", "description": "yadda yadda1","icon": "icon2.gif","title": "Number 2 title"},
 {"id": 7, "name": "yadda2", "description": "yadda yadda2","icon": "icon3.gif"},
 {"id": 10, "name": "yadda3", "description": "yadda yadda3","icon": "icon4.gif"}
]

这将影响基于set1中的对象（在此示例中为a）进行合并，（即如果在set1中发生则包括对象，如果在set1和set2中则包含合并对象，但如果仅在set2中发生则不包括），以及处理1对多关系（即set2中的对象的多个实例将与set1的obj合并为一个对象）并且对象的顺序无关紧要。你甚至可以更进一步传入两个对象中的关键字（公共元素）字段，在这个例子中，“id”被硬连线到mergeObjArrays函数。

Answer 3

http://plnkr.co/edit/77kMDeqkGvyVK87FISGl?p=info

function combine(x,y){
  var z = x;
  for (i=0; i<x.length; i++){
    for (j=0; j<y.length; j++){
      if (x[i].id === y[j].id) {
        z[i].icon = y[j].icon;
        break;
      }  
    }
  }
  return z;
}

Answer 4

似乎从SQL JOIN迁移到Javascript代码。

这是LEFT JOIN工具：

var hash2 = _.object(
    _.map(list2,function(row){ 
        return [row.id,_.omit(row,"id")]; 
    })
);
var newList = _.map(list1,function(row){ 
    return _.extend(row,hash2[row.id]||{}); 
});

Javascript基于公共元素合并数组

4 个答案: