Question

我正在寻找在JavaScript中将相关数组合并在一起的最干净，最酷的方式。

我的例子就是这样：

我从我的API获取了两个JSON数组：问题和位置。

问题有location_id，因此我想为每个问题提供一个location字段，该字段具有正确的location对象，具体取决于问题{{1} }。

如果我有这些数据：

location_id

丑陋的解决方案是：

var issues = [{id: 1, title: 'issue 1', location_id: 1}, {id: 12, title: 'issue 1', location_id: 2}];

var locations = [{id: 1, name: 'location 1'}, {id: 2, name: 'location 2'}];

生成的for(i = 0; i < issues.length; ++i) { for(j = 0; j < locations.length; ++j) { if(issues[i].location_id == locations[j].id) { issues[i].location = locations[j]; break; } } }数组将是：

issues

我正在尝试（并且失败）使用.map（）来提出更短的解决方案甚至是一个班轮。

任何指导赞赏!! ：）

Answer 1

使用map和filter：

issues.map(function (issue) {
    issue.location = locations.filter(function(location) {
      return issue.location_id === location.id;
    })[0];
    return issue;
});

Answer 2

使用object保留地图，时间复杂度为O(n+m)而不是O(n*m)。

var issues = [{id: 1, title: 'issue 1', location_id: 1}, {id: 12, title: 'issue 1', location_id: 2}];
var locations = [{id: 1, name: 'location 1'}, {id: 2, name: 'location 2'}];

var binds = function(locationList, issueList) {
  var locMap = {};
  
  // clone. If you want to directly modify the issues, this line is no need.
  issueList = JSON.parse(JSON.stringify(issueList));
  
  // construct map from location list.
  locationList.forEach(function(location) {
    locMap[location.id] = location;
  });
  
  // Use the map to bind location and issue.
  issueList.forEach(function(issue) {
    var loc = locMap[issue.location_id];
    if (loc) {
      issue.location = loc;
    }
  });
  
  
  // If you don't want to clone, this line is no need.
  return issueList;
};

var bindResult = binds(locations, issues);
console.log(bindResult);

Answer 3

您可以使用forEach和过滤器

var result = issues.forEach(function (issue) {
    issue.location = locations.filter(function (location) {
        return location.id == issue.location_id;
    })[0]
})

在console.log（结果）

时为您提供此输出

[{
    "id": 1,
    "title": "issue 1",
    "location_id": 1,
    "location": {
        "id": 1,
        "name": "location 1"
    }
}, {
    "id": 12,
    "title": "issue 1",
    "location_id": 2,
    "location": {
        "id": 2,
        "name": "location 2"
    }
}]

Answer 4

1。您可以将代码优化为：

var locationObj = {};
for(i = 0; i < location.length; i++) {
    locationObj[location[i].id] = location[i];         
}
for(i = 0; i < issues.length; i++) {
    if(issues[i].location_id){
           issues[i].location = locationObj[location_id] 
    }
}

它将缓存一个对象中的所有位置，并将直接使用该位置将其详细描述为该对象的属性。执行速度更快，而不是每次都在位置数组上使用过滤器或贴图。

2。如果您的位置的ID和位置数组的索引位于同步，那么以下将是一个更好，更快的解决方案。

for(i = 0; i < issues.length; i++) {
        if(issues[i].location_id){
           issues[i].location = locations[location_id-1] 
        }
    }

将2个相关数组映射到

4 个答案: