在php中使用html窗体中的数据

Question

<form action='main.php' method='POST'>
<select name="Category" class='listbox'>
<?php $cat = mysql_query("select cName from category");
while($drop = mysql_fetch_array($cat))
{
    echo '<option value="' . $drop['Category'] . '">' . $drop['cName'] . '</option>';
}
?>
<input type='text' name='search' class='namebox'>
<input type='submit' name='submit' value='Search' class='submitbox'></select>
</select>

​

我试图使用这个表单从我的数据库中创建一个搜索引擎，但是无法从下拉菜单中获得值。

$submit = $_POST['submit'];

if($submit)
{
    $search = $_POST['search'];
    $catval = $_POST['Category'];
    echo $catval ;
    $searchval = mysql_query("select * from item where iname like '%$search%'and cId in (select cId from category where cName = '$catval')");
    while($info = mysql_fetch_array($searchval))
    {
        echo "Item Name: " . $info['iName'];
    }
}

所以当我尝试使用这种方法搜索时，我没有得到任何结果。

Answer 1

你把选择结束标签放错了:)我想你也记得关闭你的表单？

<form action='main.php' method='POST'>
<select name="Category" class='listbox'>
<?php $cat = mysql_query("select cName, Category from category");
while($drop = mysql_fetch_array($cat))
{
    echo '<option value="' . $drop['Category'] . '">' . $drop['cName'] .'</option>';
}
?>
</select>
<input type='text' name='search' class='namebox'>
<input type='submit' name='submit' value='Search' class='submitbox'> 
</form>

Answer 2

<form action='main.php' method='POST'>
<select name="Category" class='listbox'>
<?php $cat = mysql_query("select cName from category");
while($drop = mysql_fetch_array($cat))
{
    echo '<option value="' . $drop['Category'] . '">' . $drop['cName'] .         
'</option>';
}
?>
</select>
<input type='text' name='search' class='namebox'>
<input type='submit' name='submit' value='Search' class='submitbox'>
</form>

这应该能起作用。你的</select>在错误的地方。

同时，您可以使用想要查看PDO和绑定值，而不是使用mysql()，因为它已经折旧(即不再使用)和不安全。