问题| 从db获取随机值，如何不重复它们

问题描述：

I am creating a website that gets random videos from the db. My problem is that my code to get the random video is this:

select id,content from videos order by rand() limit 1

And I don't want the user to see the same video until 3 other videos (at least) were played before.

Do you have any suggestion on how to do that? This is how my site works currently.

HTML-AJAX(CALL FOR VIDEO URL)
PHP(RETURN RANDOM VIDEO URL) One video.
AJAX(DISPLAY VIDEO)

[EDITED] Another problem I am facing is that I need to return only one video url, because this is how my ajax call look like:

success: function(data){

            $('#content').html('<div id="ytapiplayer">You need Flash player 8+ and JavaScript enabled to view this video.</div>');
            var params = { allowScriptAccess: "always" };
            var atts = { id: "ytapiplayer" };
            swfobject.embedSWF(data.vidData+"&enablejsapi=1&playerapiid=ytapiplayer?autoplay=1", "ytapiplayer", "500", "405", "8", null, null, params, atts);
}

Thanks in advance.

我正在创建一个从db获取随机视频的网站。我的问题是我的代码随机获取视频是这样的： p>

 选择id，来自视频的内容按rand（）限制1 
  code>  pre> 
 
 我 不希望用户在播放之前播放的其他3个视频（至少）之前看到相同的视频。 p> 
 
 
您对如何操作有什么建议吗？
这就是我的意思 网站目前正常工作。 p> 
 
 
 HTML-AJAX（ CALL FOR VIDEO URL  em>） li> 
 
 PHP（ 返回随机视频网址 em>）一个视频。 li> 
 
 AJAX（显示视频 em>） li> 
  ol> 
 
   [编辑] 
我面临的另一个问题是我只需要返回一个视频网址，
因为这是我的ajax调用的样子： p> 
 
 
  success：function（data  ）{
 
 $（'＃content'）。html（'＆lt; div id =“ytapiplayer”＆gt;您需要启用Flash Player 8+和JavaScript才能观看此视频。＆lt; / div＆gt;'）; \  n var params = {allowScriptAccess：“alwa  ys“}; 
 var atts = {id：”ytapiplayer“}; 
 \ swfobject.embedSWF（data.vidData +”＆amp; enablejsapi = 1＆amp; playerapiid = ytapiplayer？autoplay = 1“，”ytapiplayer“，”500“，  “405”，“8”，null，null，params，atts）; 
} 
  code>  pre> 
 
 提前致谢。 p> 
  div  >

答

Could the video id be sent to the client? Then from there the client (Javascript) request the video. Here how that could be played:

Ajax the the list of video ids
insert them in an Array in javascript (var toWatch)
Random the array
Get the first video
Remove the id from the first array. You might want to keep a trace of that id in an other array
repeat 4-5

In javascript it might look like :

$.post("getVideoId.php",function(videoId){
   var aVideoToWatch = videoId.split(',').sort(randOrd);

   for(var x=0; x<aVideoToWatch.length;x++){
       $.post("getAVideo(aVideoToWatch[x])",function(){
           //play the video
       })
   }
})

// source : http://javascript.about.com/library/blsort2.htm
function randOrd(){
 return (Math.round(Math.random())-0.5); }

答

You could just use (We'll grab fifteen so we don't have to query the server so much):

SELECT id, content FROM videos ORDER BY RAND() LIMIT 15

Hand those fifteen to the browser and allow it to request the videos. Once it has seen the fifteen, it can ask the server for three more. It can skip ones it's already played by storing played video ids.

If you are delivering results via Ajax and JSON you can just return the concatenation of the results arrays:

<?php
$query = $pdo->query('SELECT id, content FROM videos ORDER BY RAND() LIMIT 10');
$videos = $query->fetchAll();
echo json_encode($videos);
?>

Then in JS:

(function playRandomVideos() {  //Closures are cool
    $.getJSON('getRandomVideos.php', {success: function(videos) {
        for(video in videos) {
            if(video.id in playRandomVideos.played) {
                continue;
            }
            play(video);
            playRandomVideos.played.push(video.id);
        }
        playRandomVideos();
    }});
})();
playedRandomVideos.played = [];

问题| 从db获取随机值，如何不重复它们

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