基於公共元素的Javascript合併數組

Question

我試圖創建一個單獨的對象，它從兩個單獨的對象（從Ajax調用）獲取信息。基本上我們有一個標籤列表和一個元素列表，這些項目是相關的，但是標籤對象並不包含我需要的所有信息來填充所有的數據。 例如：

的標籤列表：

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

我需要的ID兩個比較對象，並把它們合併成一個：

[ 
    {"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（詳細信息）的列表這個新對象使我可以訪問這兩個對象中的所有數據。

如果有幫助，這是一個角度項目，我已經在使用Underscore，我相信必須有一些方法可以用下劃線來做到這一點，但我並不是非常熟悉它。

Answer 1

快速和骯髒的：

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/

正如你所看到的，這個假設有一個1：在兩個數組對象之間一對一的關係。

Answer 2

對於什麼它的價值，沒有圖書館，只是香草JS

函數，它的工作：

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; 
    } 
} 

用法示例

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"} 
] 

這將ineffect做基於對象合併在set1中（在這個例子中是a）（即，如果在set1中包含對象，或者如果在set1和set2中都包含合併對象，而不是如果它僅在set2中出現），並且處理1對多關係作爲好（即set2中的一個對象的多個實例將被合併到一個具有set1的obj的對象中），並且對象的順序無關緊要。你甚至可以更進一步，傳入兩個對象中的關鍵字（通用元素），在本例中，「id」被硬連線到mergeObjArrays函數。

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

好像從SQL遷移到JOIN Javascript代碼。

這裏有一個LEFT JOIN實現：

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]||{}); 
});