如何在单独的方法中加载json

Question

我需要动态生成播放列表。提取播放列表的最佳方法是什么，以便将下面的播放列表 json 分成不同的方法或变量。

Mypage.foo

            <script type='text/javascript'>      
                var videoid = '1413';
                var videoUrl = 'myflv.flv';
            </script> 

## /scripts/flowplayer/init.js

$f(videoid, "/swf/flowplayer-3.1.5.swf", {

    playlist: [
        {
            url: videoUrl
        },
        {
            url: '59483.flv',
            title: 'Palm trees and the sun'
        },
        {
            url: '58192.flv',
            title: 'Happy feet in a car'
        },
        {
            url: '63617.flv',
            title: 'People jogging'
        }
    ],
    wmode: 'opaque',
    plugins: {
        gatracker: {
            url: "/swf/flowplayer.analytics-3.1.5.swf",
            trackingMode: "Bridge",
            debug: false
        }
    }
}); 

Answer 1

我想你想要这样的东西：

  var list1 = getPlayList();

  $f(videoid, "../flowplayer-3.1.5.swf", {
      playlist: list1,
      wmode: 'opaque'
  });

...其中 getPlayList() 返回一个按照你需要的方式格式化的数组：

  function getPlayList() {
      return [
      {
          title: 'scooter race',
          url: 'http://www.mediacollege.com/video-gallery/testclips/20051210-w50s.flv'
      },
      {
          title: 'Promo',
          url: 'http://e1h13.simplecdn.net/flowplayer/flowplayer.flv'
      }
      ];
  }

Answer 2

只需提取它：

var f=function(param1,param2,...){
  return [
    // dynamic playlist data
  ];
}

然后调用它：

$f(videoid, "/swf/flowplayer-3.1.5.swf", {
  playlist: function(){
    return f(param1,param2,...);
  }
});