Php mysql数据的下拉菜单
我有以下代码:Php mysql数据的下拉菜单,php,html,dropdown,Php,Html,Dropdown,我有以下代码: <?php session_start(); include_once 'include/dbconnect.php'; if(!isset($_SESSION['user'])) { header("Location: ../index.php"); } $dropdown = mysql_query("SELECT * FROM carros_real WHERE real_dono='".$_SESSION['user']."'"); while ($drop_ro
<?php
session_start();
include_once 'include/dbconnect.php';
if(!isset($_SESSION['user']))
{
header("Location: ../index.php");
}
$dropdown = mysql_query("SELECT * FROM carros_real WHERE real_dono='".$_SESSION['user']."'");
while ($drop_row = mysql_fetch_array($dropdown)){
echo "<select>";
echo "<option value='".$drop_row['real_id']."'>" . $drop_row['real_marca'] . "</option>";
echo "</select>";
}
?>
将您的
和
移出您的循环,同时感谢您。。。但是你能举个例子吗?谢谢!!向上投票!这工作做得很好。现在我可以将其插入表单并获得$\u POST['name\u here'],对吗?如果您在某些地方出错,这可能会导致SQL注入漏洞。为了安全起见,请转义$SESSION['user']
。或者更好的方法是,切换到mysqli\uquot>或PDO,并使用准备好的语句!:-)还有一个问题。。。如果($_GET['option']){}我试图插入一个name='car'并将其分配给do$var=$_GET['car'],如何将所选选项值分配给$var以使用;但这并不奏效。谢谢
<?php
session_start();
include_once 'include/dbconnect.php';
if(!isset($_SESSION['user']))
{
header("Location: ../index.php");
}
$dropdown = mysql_query("SELECT * FROM carros_real WHERE real_dono='".$_SESSION['user']."'");
echo "<select>";
while ($drop_row = mysql_fetch_array($dropdown)){
echo "<option value='".$drop_row['real_id']."'>" . $drop_row['real_marca'] . "</option>";
}
echo "</select>";
?>