如果name有超过1个单词,则mysql查询失败

Question

我开始为我的小型房地产业务创建一个网站.我玩了一些函数http://www.php.net mysql,我设法通过AJAX访问页面并返回搜索引擎的html内容.

我有一个已经填充了公寓和房屋的数据库

问题是如果公寓名称是"公寓"我返回html内容,如果"公寓有3个房间"它不再写任何东西.

我不明白我错在哪里:

<?php
    $search = $_GET['selected'];

    $link = mysql_connect('localhost', 'root', '');
    mysql_select_db('houses', $link);

    function searchHouse($search, $link){
        $query = "select * from houses where name=$search limit 1";
        $result = mysql_query($query);
        $row = mysql_fetch_assoc($result);

        $query2 = "select * from houses_info where house_id=$row[id]";
        $result2 = mysql_query($query2);
        $row = mysql_fetch_assoc($result2);
        return $row;
    }

    $result = searchHouse($search, $link);

    echo $result['house_sq'];
    echo "<br>";
    echo $result['house_rooms'];
    echo "<br>";
    echo $result['house_bathrooms'];
    echo "<br>";
    echo $result['house_address'];

    ?>

Answer 1

• 你应该知道你是否"玩过" php.net,这些mysql_*功能已弃用且不再维护.它是页面顶部的红色框,通知您.
• 你有一个大的MySQL喷射孔那里,你没有逃避$string可言
• 你的问题是你没有添加引号,$string如:'$string'
• 你应该使用stat PDO来摆脱坏的代码和SQL注入漏洞.

• 你可以将这2个选项包装成一个选择:

  <?php
      function searchHouse($search, $link){
          $search = mysql_real_escape_string($search);
          $query = "select * from houses_info where house_id IN (select * from houses where name='".$search."' limit 1)";
          $result = mysql_query($query);
          $row = mysql_fetch_assoc($result);
          return $row;
      }
  ?>
  

• 既然你已经开始建立那个网站了,你可以开始学习PDO,阅读本教程,你的代码会更像这样:

  <?php
      $db = new PDO('mysql:host=localhost;dbname=houses;charset=UTF-8', 'root', '', array(PDO::ATTR_EMULATE_PREPARES => false, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));
      $search = $_GET['selected'];
  
      function searchHouse($search){
          global $db;
  
          $query = $db->prepare("select * from houses_info where house_id IN (select * from houses where name=:search limit 1)");
          $query->execute(array(':search' => $search));
          $row = $query->fetch(PDO::FETCH_ASSOC);
          return $row;
      }
  
      $result = searchHouse($search);
  ?>