Question

我正在尝试使用ajax（在该区域非常新手）和/或jquery用我的mysql查询结果刷新div。我的index.php看起来像这样：

<head>
$( document ).ready(function() {
(function showscreen() {
   $.ajax({
    url: 'screen.php', 
    success: function(data) {
       $('#screen').html(data);
    },
    complete: function() {

      setTimeout(showscreen, 5000);
    }
  });
})(); 
});
</head>
<body>
<div id="screen"></div>
<?php

require '../dbauth/videos.inc.php';
//This handles my db connections.

$videosdbhandle = mysql_connect(

$videoshostname, 
$videosusername, 
$videospassword

) or die("INVALID USERNAME OR PASSWORD");

$videosselected = mysql_select_db($videosdbname, $videosdbhandle);

$checkvideosql = "SELECT * FROM videos ORDER BY votes DESC";    
$checkvideoquery = mysql_query($checkvideosql, $videosdbhandle); 
$checkvideocount = mysql_num_rows($checkvideoquery); 

if ($checkvideocount > 0) {

$count = 0;

while ($checkvideocount > $count) {

$thevideoname = mysql_result($checkvideoquery, $count, "name");
$thevideovideo = mysql_result($checkvideoquery, $count, "video"); 
$thevideoid = mysql_result($checkvideoquery, $count, "id"); 
$thevideoimage = mysql_result($checkvideoquery, $count, "image"); 

    echo '<div class="screen" id="video';
    echo $thevideoid; 
    echo '" style="display: none; background-image: url(';
    echo "'http://backcurrents.com/videos/";
    echo $thevideoname;
    echo '/';
    echo $thevideoimage;
    echo "'";
    echo ');">';

    echo '<video loop muted autoplay 
poster="http://backcurrents.com/videos/';
    echo $thevideoname;
    echo '/';
    echo $thevideovideo; 
    echo '" class="videobg">
    <source src="http://backcurrents.com/videos/';
    echo $thevideoname;
    echo '/';
    echo $thevideovideo; 
    echo '" type="video/mp4">
    </video>';
    echo '</div>';


$count++;
}


}else{

echo 'NO VIDEOS!';

}

?>
</body>

然后我的screen.php看起来像这样：

$videosusername = "theusername";
$videospassword = "thepassword";
$videoshostname = "localhost";
$videosdbname = "thedbname";

$videosdbhandle = mysql_connect(

$videoshostname,
$videosusername, 
$videospassword

) or die("INVALID SHOW NAME OR PASSWORD");

$videosselected = mysql_select_db($videosdbname, $videosdbhandle);

$checkhighestvideosql = "SELECT * FROM videos ORDER BY votes DESC";
$checkhighestvideoquery = mysql_query($checkhighestvideosql, 
$videosdbhandle);
$checkhighestvideocount = mysql_num_rows($checkhighestvideoquery); 



if ($checkhighestvideocount > 0) {
$x=0; 

while ($checkhighestvideocount > $x) {

$checkhighestvideoid = mysql_result($checkhighestvideoquery, $x, "id"); 

if ($x=0) {

echo '<script type="text/javascript">';
echo '$(".screen").hide();
$("#video';
echo $checkhighestvideoid; 
echo '").show();  ';
echo '</script>';

}else{
echo '<script type="text/javascript">';
echo '$(".screen").hide();
';
echo '</script>';
}

$x++;
 }

}else{
echo "NO VIDEOS!";  
}

我想要做的就是每隔5秒刷新一次mysql查询，但是屏幕中没有任何内容出现在screen.php中。我已经尝试了几次不同的迭代，错误检查我的ajax代码是通过简单地放置：echo“TEST TEXT”;在我的screen.php中，这是有效的。任何时候我将一个mysql查询插入screen.php，什么都没有出现。有没有比这更好的方法来实现我想要的？我基本上有一个投票系统内置到一个存储投票的mysql数据库中。投票最多的视频会在我建立的index.php文件中播放。

定期刷新MySql查询

0 个答案: