我对下面的代码有问题，它显示的是html部分，而不是我数据库中的数据。我只是想把数据库中的数据放在下拉列表中，然后单击find将其显示在下表中。你能帮忙吗？
<?php
include 'config.php';
?>
<html>
<head>
<meta charset="UTF-8">
<title></title>
</head>
<body>
<select name="manufacturer">
<option value="">----ALL----</option>
<?php
$sql = "SELECT id, manufacturer FROM coop";
$result = $conn->query($sql);
while ($r = mysqli_fetch_assoc($result))
{
echo "<option value='$r[id]'> $r[manufacturer]</option>";
}
?>
</select>
<input type="submit" name="find" value="find"/>
<br><br>
<table border="1">
<tr align="center">
<th>ID</th> <th>Manufacturer</th>
</tr>
<?php
if($_SERVER['REQUEST_METHOD'] == "POST")
{
$des=$_POST["manufacturer"];
if($des=="")
{
$result= mysqli_query("SELECT id, manufacturer FROM coop");
}
else
{
$result= mysqli_query($sql);
}
echo "<tr><td colspan='5'></td></tr>";
while($r= mysqli_fetch_assoc($result))
{
echo "<tr>";
echo "<td align='center'>$r[0]</td>";
echo "<td width='200'>$r[1]" . " $r[2]</td>";
echo "<td alig='center' width='40'> $r[3]</td>";
echo "<td align='center' width='200'>$r[4]</td>";
echo "<td width='100' align='center'>$r[5]</td>";
echo "</tr>";
}
}
?>
</table>
</body>
</html>
此处的queries错误
<?php
$sql = "SELECT id, manufacturer FROM coop";
$result = $conn->query($sql);
while ($r = mysqli_fetch_assoc($sql))
{
echo "<option value='$r[0]'> $r[0]</option>";
}
?>
这个while ($r = mysqli_fetch_assoc($sql))应该是while ($r = mysqli_fetch_assoc($result ))
你应该用$result而不是$sql 获取你的assoc
在部分下面仍然有一些错误