Question

您好试图从wordpress媒体播放列表中请求每个src值，因此我以JSON格式返回数据

我的请求未成功的代码是：

$decode_tracks = json_decode($track , true);

foreach ($decode_tracks[0][0] as $item) {
 print $item['src'];
}

这是返回的一些数据。

[
      {
        "src": "http:\/\/localhost\/wordpress\/wp-content\/uploads\/2015\/08\/output.mkv",
        "type": "video\/x-matroska",
        "title": "output",
        "caption": "",
        "description": "",
        "meta": {
          "length_formatted": "0:16"
        }
      },
      {
        "src": "http:\/\/localhost\/wordpress\/wp-content\/uploads\/2015\/08\/christmas.mp4",
        "type": "video\/mp4",
        "title": "christmas",
        "caption": "tytj",
        "description": "tyjtyj",
        "meta": {
          "length_formatted": "1:21"
        }
      }
    ]

Answer 1

试试这个：

foreach ($decode_tracks as $item) {
     print $item['src'];
}

无需将json转换为数组，您可以轻松地执行此操作：

foreach ($track as $item) {
     print $item->src;
}

Answer 2

$decode_tracks = json_decode($track);

for ($i = 0; $i < count($decode_tracks); $i++) {
  print $decode_tracks[$i]->src;
}

或者如果您正在使用数组：试试这个

$decode_tracks = json_decode($track, true);

for ($i = 0; $i < count($decode_tracks); $i++) {
  print $decode_tracks[$i]['src'];
}

在PHP中解析JSON对象并使用json_decode获取每个值

2 个答案: