php - 循环和回显指定的 JSON 内容

Question

如何仅在" wrapperType ":"track"中显示艺术家姓名？

这是 JSON 示例：

{  
   "resultCount":5,
   "results":[  
      {  
         "wrapperType":"collection",
         "artistName":"Liam Payne"
      },
      {  
         "wrapperType":"track",
         "artistName":"Liam Payne & French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"Liam Payne & French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"French Montana"
      }
   ]
}

和php代码：

但这将显示所有值

foreach ($obj->results as $row){
    echo $row->artistName;
}

标签： phparraysjson

Answer 1

将if($row->wrapperType == "track")条件放入您的代码中。

试试下面的代码：

foreach ($obj->results as $row){
    if($row->wrapperType == "track"){
        echo $row->artistName;
    }
}