从菜单中显示 MYSQL 数据
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.
相关文章