来自 mysql 数据的下拉菜单

Question

我有这个代码：

<?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 ($drop_row = mysql_fetch_array($dropdown)){
echo "<select>";
echo "<option value='".$drop_row['real_id']."'>" . $drop_row['real_marca'] .     "</option>";
echo "</select>";
}
?>

结果是这样的： http://prntscr.com/9d1vpg

嗯，我打算做的恰恰相反……我需要一个包含所有汽车的列表，我想将 $drop_row['real_id'] 发送到 $var。我在这里缺少什么？谢谢

Answer 1

<?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>";
 ?>

来自 mysql 数据的下拉菜单

Dropdown menu from mysql data

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php

dropdown