Question

我想使用PHP从数据库中播放我的视频，因为我将video标记放在我的echo语句中，视频无法播放。请告诉我我做错了什么

<?php
    $query1 = mysql_query("select * from video_title where id='541' and status=1"); 
    while($qry1 = mysql_fetch_array($query1)) { 
        $vid = $qry1['video'];    ?>  


    <video style="border:1px solid" id="myVideo" width="320" height="176" controls >
      <source src="<?php echo $vid;?>" type="video/mp4">
    </video> 
    <br>

    <button onclick="slowPlaySpeed()" type="button"><img src="../user/img/download.jpg" title="Slow Play" height="30px"></button>
    <button style="margin-left:45px" onclick="nrmlPlaySpeed()" type="button"> play normal</button>
    <button style="margin-left:45px" onclick="fastPlaySpeed()" type="button"><img src="../user/img/arrowRight.gif" title="Fast Play" height="30px"> </button>

    <script>
    // Get the video element with id="myVideo"
    var vid = document.getElementById("myVideo");

    // Set the current playback speed of the video to 0.3 (slow motion)
    function slowPlaySpeed() { 
        vid.playbackRate = 0.3;
    }
    function nrmlPlaySpeed() { 
        vid.playbackRate = 1.0;
    }
    function fastPlaySpeed() { 
        vid.playbackRate = 5.0;
    }

    // Assign an onratechange event to the video element, and execute a function if the playing speed of the video is changed

    </script>
<? }?>

Answer 1

//mysql query 

$sql = "select * from video_title where id='541' and status=1";
$result = mysql_query($conn, $sql); //$conn should be db query

while($row = mysql_fetch_array($result , MYSQL_ASSOC)){
    $vid=$row['video'];
?>
<video width="100%" controls>
  <source src="<?php echo $vid ?>;" type="video/mp4">
</video>
<?php
}

Answer 2

您应将html和javascript代码置于while循环之外您应该使用mysqli_query($conn, $sql)代替mysql_query($sql)。

自PHP 5.5.0起，所有mysql_functions都已弃用。因此不再使用它们。

<?php
$query1 = mysqli_query($conn, "select * from video_title where id='541' and status=1");
while ($qry1 = mysqli_fetch_assoc($query1)) {
    $vid = $qry1["video"];
}
?>

<video width="100%" controls>
    <source src="<?php echo $vid ?>;" type="video/mp4">
</video>
<br>
<button onclick="slowPlaySpeed()" type="button"><img src="../user/img/download.jpg" title="Slow Play" height="30px"></button>
<button style="margin-left:45px" onclick="nrmlPlaySpeed()" type="button"> play normal</button>
<button style="margin-left:45px" onclick="fastPlaySpeed()" type="button"><img src="../user/img/arrowRight.gif" title="Fast Play" height="30px"></button>

<script>
    var vid = document.getElementById("myVideo");

    function slowPlaySpeed() {
        vid.playbackRate = 0.3;
    }
    function nrmlPlaySpeed() {
        vid.playbackRate = 1.0;
    }
    function fastPlaySpeed() {
        vid.playbackRate = 5.0;
    }
</script>

如何使用PHP从数据库获取播放视频

2 个答案: