我希望我的下拉列表在选项中显示来自sql表的名称,但将值作为sql表中的ID。我如何实现这一目标?将HTML下拉列表中每个选项的值设置为一个PHP变量
<select name="Warehouse">
<?php
$servername = "127.0.0.1";
$username = "root";
$password = "12345";
$db_name = "Second";
$the_port = "3306";
// Create connection
$conn = new mysqli($servername, $username, $password,$db_name,$the_port);
$conn->set_charset('utf8');
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
echo "err";
}
// fetch ID,Name_of from table
$sql="SELECT ID,Name_of FROM WAREHOUSE;";
$result = $conn->query($sql);
while($row = mysqli_fetch_assoc($result)) {
$val=$row['ID'];
echo "<option value=".$val." >" . $row['Name_of'] . "</option>";
}
?></select>
那么是什么问题? – 2015-03-25 06:07:26
接受任何人回答,以便其他人知道 – 2015-03-25 06:37:59
我已经回答了... – 2015-03-26 08:25:39