Javascript基于公共元素合并数组

我正在尝试创建一个单个对象，该对象从两个单独的对象（取自ajax调用）获取信息。 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做这个，但我不是很熟悉它。

Answer 1

快而脏：

 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

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

 function combine(x,y){ var z = x; for (i=0; i 

Answer 3

似乎从SQL JOIN迁移到Javascript代码。

这是一个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]||{}); }); 

Answer 4

对于它的价值，没有图书馆，只有香草js

完成工作的function：

 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"} ] 

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），并处理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。 您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，在此示例中，“id”硬连线到mergeObjArrays函数。