Display MYSQL Data From a Menu(从菜单中显示 MYSQL 数据)

本文介绍了从菜单中显示 MYSQL 数据的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个 MYSQL 数据库，其中包含许多字段，例如财产、卧室、大小等

I have a MYSQL database with a number of fields such as property, bedrooms, size etc

我有两个下拉列表，其中包含数据库中的数据

I have two dropdown list with data that is contained within the database

提交选项时，我希望打开一个显示结果的新页面.我收到错误消息 mysql_fetch_assoc(): 提供的参数不是有效的 MySQL 并且不知道如何解决这个问题！非常感谢帮助...我了解 SQL 注入，并希望在我让本节开始工作后纠正此问题

When submitting the options I want a new page to open displaying the results. I am getting the error message mysql_fetch_assoc(): supplied argument is not a valid MySQL and have no idea how to fix this! help much appreciated...I know about SQL injections and looking to rectify this after I get this section working first

HTML

     <form method="get" action="submit.php">

     Number:  <select name="property">
     <option value="Aviemore House">Aviemore House</option>
     <option value="Dalfaber House">Dalfaber House</option>
     </select>
     <br>

     Name: <select name="bedrooms">
     <option value="2">2</option>
     <option value="3">3</option></select>
     <br>

    <input type="submit" value="submit" />
    </form>

PHP

 <?php

 require 'defaults.php';
 require 'database.php';


 $property = $_GET['property'] ;
 $bedrooms = $_GET['bedrooms'] ;

 $query = "select FROM properties where property = '$property' & bedrooms = '$bedrooms'";

 while ($row = mysql_fetch_assoc($result))
 {
$r[] = $row;
 }

 ?>

推荐答案

试试这个:

$query = "SELECT * FROM `properties` WHERE property = '{$property}' AND bedrooms = '{$bedrooms}'";
$row=mysql_query($query);

您的 sql 格式错误，需要执行查询.

Your sql is malformatted and need to execute the query.

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