我可以通过索引而不是名称来引用ngRepeat中的项目(Can I refer items inside ngRepeat by index, not name)

我可以通过索引而不是名称来引用ngRepeat中的项目(Can I refer items inside ngRepeat by index, not name)

我要找的是取代这个：

<tr ng-repeat="travel in travelList"> <td> {{travel.travelerId}} </td> <td> {{travel.travelerName}} </td> </tr>

有类似的东西：

<tr ng-repeat="travel in travelList"> <td> {{travel[0]}} </td> <td> {{travel[1]}} </td> </tr>

TravelList在TravelList中的数据结构如下：

public class Travel { public int TravelId { get; set; } public string TravelerId { get; set; } public string TravelerName { get; set; } public string TravelerEmail { get; set; } public string RecruiterId { get; set; } public List<Itinerary> Itinerary { get; set; } }

请指教。

What I am looking for, is to replace this:

<tr ng-repeat="travel in travelList"> <td> {{travel.travelerId}} </td> <td> {{travel.travelerName}} </td> </tr>

With something like:

<tr ng-repeat="travel in travelList"> <td> {{travel[0]}} </td> <td> {{travel[1]}} </td> </tr>

The data stracture for Travel inside TravelList is this:

public class Travel { public int TravelId { get; set; } public string TravelerId { get; set; } public string TravelerName { get; set; } public string TravelerEmail { get; set; } public string RecruiterId { get; set; } public List<Itinerary> Itinerary { get; set; } }

Please advise.

最满意答案

不，这不是JS对象的工作方式。 someObj[key]是有效的语法，但除非你明确设置一个键的值（如someObj[0] = 'foo' ），否则结果是未定义的。

如果你想迭代每次travel的属性，你需要<td ng-repeat='(key, value) in travel'>{{travel[key]}}</td>进行第二次ng-repeat例如<td ng-repeat='(key, value) in travel'>{{travel[key]}}</td> ，但是你可能只使用{{value}}除非你希望在标题行或类似的东西中使用键作为标签。

No. That's not how JS objects work. someObj[key] is valid syntax, but unless you explicitly set a value for a key (like someObj[0] = 'foo') then the result is undefined.

If you want to iterate over the properties of each travel, you will need a second ng-repeat like <td ng-repeat='(key, value) in travel'>{{travel[key]}}</td>, but then you might as well just use {{value}} unless you are hoping to use the keys as labels in a header row or something like that.