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所以我有一些php代码来将一个mysql表返回到一个下拉框中，如下所示：

$hostname = ""; //SET SERVER/HOSTNAME
$dbusername = ""; //SET DATABASE USERNAME
$dbname = ""; //SET DATABASE NAME
$dbpassword = ""; //SET DATABASE USERNAME
$link = mysqli_connect($hostname, $dbusername, $dbpassword, $dbname); 

if (!$link)
{ 
    die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error()); 
} 

$sql = "SELECT * FROM utility ORDER BY program_code"; 

if ($result = mysqli_query($link, $sql)) 
{    
    while ($row = mysqli_fetch_assoc($result)) 
    {
        $selectbox.='<option value=\"' . $row['program_code'] . '\">' 
        . $row['program_code'] . ' - ' . $row['rate'] . ' - ' . $row['term'] . '</option>';
    }

    $selectbox.='</select>';

    mysqli_free_result($result);

    echo $selectbox;
}

这很好用，然后我就把它加入我的表格中。My problem is when an option is selected, I need to get and return that value so I can pass it on to whatever else I am doing with the values from my form. 任何帮助是极大的赞赏。谢谢你。

4

1 回答 1

1 
<select name="select">
  <option value="foo">foo</option>
</select>

PHP：

$select = $_POST["select"];
echo $select; //Should return foo
于 2013-08-02T18:48:14.087 回答