2016-02-13 129 views
0

我刚开始查看一个旧项目,并且遇到了一个问题。 我想在表格中显示一些数据,但我忘了它是如何与json对象和角度一起工作的。在带有角度的表格中显示json数据

我从一个看起来像这样的API获取数据。

{"Search":[{"Title":"Not Another Teen Movie","Year":"2001","imdbID":"tt0277371","Type":"movie","Poster":"http://ia.media-imdb.com/images/M/[email protected]_V1_SX300.jpg"},{"Title":"Not Just Another 8 Teen Movie","Year":"2003","imdbID":"tt0381457","Type":"movie","Poster":"N/A"},{"Title":"Not Just Another 8 Teen Movie 2","Year":"2003","imdbID":"tt0397579","Type":"movie","Poster":"N/A"},{"Title":"Not Just Another 8 Teen Movie 3","Year":"2004","imdbID":"tt0408045","Type":"movie","Poster":"N/A"}],"totalResults":"4","Response":"True"} 

那么我作用域的数据是这样

$scope.AllMoviesFound = data; 

然后我想告诉所有的不同的称谓在我的表

<table class="table table-striped table-hover"> 
      <thead> 
      <tr class="success"> 
       <th> 
         <h3>Title</h3></a> 
       </th> 
      </tr> 
      </thead> 
      <tbody> 
      <tr ng-repeat="item in AllMoviesFound"> 
       <td><h4>{{item.Title}}</h4></td> 
      </tr> 
      </tbody> 
     </table> 

但我不能完全得到这工作。我在桌子上丢失了什么?

+1

你没有关闭ng-repeat结尾的双引号 – mariovalens

+0

谢谢,但这只是复制粘贴到stackoverflow后的错字。不是真正的问题。 ;) – Englund0110

回答

0

您可以使用

var response = '{"Search":[{"Title":"Not Another Teen Movie","Year":"2001","imdbID":"tt0277371","Type":"movie","Poster":"http://ia.media-imdb.com/images/M/[email protected]_V1_SX300.jpg"},{"Title":"Not Just Another 8 Teen Movie","Year":"2003","imdbID":"tt0381457","Type":"movie","Poster":"N/A"},{"Title":"Not Just Another 8 Teen Movie 2","Year":"2003","imdbID":"tt0397579","Type":"movie","Poster":"N/A"},{"Title":"Not Just Another 8 Teen Movie 3","Year":"2004","imdbID":"tt0408045","Type":"movie","Poster":"N/A"}],"totalResults":"4","Response":"True"}' 
; 
var JsonObject= JSON.parse(response); 

// OR

$(jQuery.parseJSON(JSON.stringify('your string')['search'])).each(function() { 
      var title = this.title; 
      //do stuff 
    }); 
-1

写作$scope.AllMoviesFound = data;意味着,例如,你有你试吃包糖果袋,不会工作。

这里你的电影数据是在一个名为数据的'盒'中。所以你必须在这个盒子里拍电影来处理它。

$scope.AllMoviesFound = data.Search 

表示您从数据中搜索并将其放入您的范围。你现在可以迭代它。

此外,您不能使用ng-repeat与对象一样,它仅适用于数组。

+1

对于ng-repeat和对象是不正确的。阅读文档 – charlietfl

+0

当我说“_Moreover不能使用ng-repeat **作为它与对象_”时,我可能并不清楚。我的坏...英语不是我的母语。无论如何,它看起来你不能用语法012er ng-repeat =“对象中的项目”' 正确的语法是'ng-repeat =“(key,value)in Object”' – Bruspal

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