Question

我试图从这个JSON中抓取artistId

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<table>
  <tr>
    <td>ClientName</td>
    <td>ItemName</td>
    <td>Quantity</td>
    <td>12</td>
    <td>34</td>
  </tr>
  <tr>
    <td>ClientName</td>
    <td>ItemName</td>
    <td>Quantity</td>
    <td>56</td>
    <td>78</td>
  </tr>
  <tr>
    <td>ClientName</td>
    <td>ItemName</td>
    <td>Quantity</td>
    <td>90</td>
    <td>12</td>
  </tr>
  <tr>
    <td colspan="3">Totals</td>
    <td id="subtotal"></td>
    <td id="total">12</td>
  </tr>
</table>

创建播客对象时。但是当我尝试将此变量显示为字符串时，它返回null。我该如何解决这个问题？

这是我尝试访问artistID的地方

{
  "wrapperType": "track",
  "kind": "podcast",
  "artistId": 125443881,
  "collectionId": 523121474,
  "trackId": 523121474,

字符串使＆＃34; podcastID＆＃34; （null）在所有情况下。

这是iTunes响应，我在其中分配了艺术家ID

  NSNumber *podcastID = episode.podcast.artistID;
NSString *episodeNumber = [episode.title stringByReplacingOccurrencesOfString:@" " withString:@"-"];
NSString *deeplinkURL = [NSString stringWithFormat: @"podcast://%@/%@",podcastID, episodeNumber];

Answer 1

尝试使用NSNumber：

self.artistId = [(NSNumber *)[response valueForKey:@"artistId"] integerValue];

NSInteger vs NSNumber并从itunes检索JSON数据

1 个答案: