<?php
mysqli_select_db($con,'media');
$query = mysqli_query($con,"select * from sounds");
while($row = mysqli_fetch_assoc($query))
{
$soundname = $row['name'];
$soundurl = $row['url'];
echo "
<h5>$soundname</h6>
<audio src=\"$soundurl\"></audio>
<form>
<select name=\"playlist\" onchange=\"this.form.submit()\">
";
mysqli_select_db($con,'playlists');
$query1 = mysqli_query($con,"select * from pl");
while($row1 = mysqli_fetch_assoc($query1))
{
$playlistname = $row1['name'];
$playlistid = $row1['id'];
echo"
<option value=\"$playlistid\">$playlistname</option>
";
}
echo"
</select>
</form>
";
if(isset($_POST["playlist"])){
mysqli_select_db($con,'added');
$que="INSERT INTO user values('','$soundname','$soundurl','$playlistid')";
$que_run = mysqli_query($con,$que);
}
}
?>
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最佳答案
也许这里有错误!
$soundname = $row['name'];
$soundurl = $row['url'];
echo "
<h5>$soundname</h5>
<audio src=\"$soundurl\"></audio>
关于php - 为什么此代码不适用于所有声音和所有播放列表?它仅适用于第一个声音和第一个播放列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55271428/