php - 此后如何使用 explode ，因为在我的数据库列中，所有值都在产品列中可用，如下所示

Question

我正在创建一个数据库来存储用户信息和链接到该用户的产品。我通过使用 , 内爆来存储产品。稍后如何 explode 它们？

HTML 表单:

<form name="frm" method="post">

<table align="center">
<tr>
    <td>Name</td>
    <td><input type="text" name="txt" id="txt" required ></td><td></td>
</tr>
<tr>
    <td>Email</td>
    <td><input type="email" name="email" id="email" required ></td><td></td>
</tr>
<tr>
    <td>Password</td>
    <td><input type="password" name="pass" id="pass" required></td><td></td>
</tr>
<tr>
    <td>Product</td>
    <td>
        <input type="checkbox" name="chk[]" value="laptop">Laptop
        <input type="checkbox" name="chk[]" value="pc">Pc
        <input type="checkbox" name="chk[]" value="tablet">Tablet
    </td><td></td>
</tr>
<tr>
    <td></td>
    <td><input type="submit" name="submit" value="Save"></td><td></td>
</tr>
</table>
</form>

在 PHP 代码中:

if(isset($_POST['submit']))
{
    $name=$_POST['txt'];
    $email=$_POST['email'];
    $pass=$_POST['pass'];
    $product=$_POST['chk'];

    foreach($product as $result)
    {
        $csk[]=$result;
    }
    //print_r($csk); exit;
    $rko=implode(",",$csk);

    $rec=mysql_query("insert into test(name,email,pass,product) values('".$name."','".$email."','".$pass."','".$rko."')");

    if($rec)
    {
        echo "Successfully Added";
    }
    else
    {
        echo " Failed To Add";
    }
}

数据库将包含如下记录:

+----+------+-------------+------+------------------+
| id | name | email       | pass | product          |
|  1 | kodi | k@gmail.com | kodi | laptop,pc        |
|  2 | muru | m@gmail.com | kodi | laptop,pc,tablet |
+----+------+-------------+------+------------------+

我想知道如何分解这个，我想知道如何编写查询来查找拥有笔记本电脑、个人电脑、平板电脑或某种组合的人。

Answer 1

您可以使用 SQL 通配符在产品字段中搜索:

SELECT * FROM `dbtable` WHERE `product` LIKE '%tablet%' OR `product` LIKE '%laptop%';

要分离结果，您可以这样做:

$res = explode(",", $result['product']);
foreach ($res as $item) {
    echo "has $item ";
}

更好的方法是:

• 您应该考虑不存储纯文本密码。

• 您应该连接表，这些表实际上具有人员和项目之间的关系。例如:

  表人员

  +----+------+-------------+------------------------------------------------------------------+
  | id | name | email       | password_hash                                                    |
  |  1 | kodi | k@gmail.com | 2cf24dba5fb0a30e26e83b2ac5b9e29e1b161e5c1fa7425e73043362938b9824 |
  |  2 | muru | m@gmail.com | c0e81794384491161f1777c232bc6bd9ec38f616560b120fda8e90f383853542 |
  +----+------+-------------+------------------------------------------------------------------+
  

  表东西

  +----+-----------+--------+
  | id | person_id | item   |
  |  1 |         1 | laptop |
  |  2 |         1 | pc     |
  |  3 |         2 | laptop |
  |  4 |         2 | pc     |
  |  5 |         2 | tablet |
  +----+------+-------------+
  

  那么您的查询将是:

  SELECT p.* 
  FROM `people` AS `p` LEFT JOIN 
       `stuff` AS `s` ON `p`.`id` = `s`.`person_id` 
  WHERE `s`.item IN ('laptop','tablet');