这里有点混乱,例如我有这样的用户表:
--------------------------------
| id | name | pass | email |
--------------------------------
| 1 | a | a | a@a.a |
--------------------------------
| 2 | b | b | b@b.b |
--------------------------------
| 3 | c | c | c@c.c |
--------------------------------
我可以使用这个命令
$query = mysqli_query($con, "SELECT * FROM user");
$row = mysqli_fetch_array($query, MYSQLI_ASSOC);
$username = $row['name'];
$password = $row['pass'];
但是当我有这个这样的选项表时会遇到问题
------------------------------
| id | name | content |
------------------------------
| 1 | username | a |
------------------------------
| 2 | password | a |
------------------------------
| 3 | email | a@a.a |
------------------------------
| 4 | url | a.com |
------------------------------
如何查询此表?重点是我想创建选项表,但不是将数据添加到右边,而是添加到底部,这样点"名称值=内容值"
username = a;
password = a;
email = a@a.a;
url = a.com;
我真的不知道如何解释,这对我来说是新的
您可以在PHP中试用变量。例如,在上面的例子中,尝试这样的方法:
$username = '';
// query your OPTIONS table here
// assuming you've called it 'user' as described in your question
$query = mysqli_query($con, "SELECT * FROM user");
while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
// note the extra $ prefix before $row['name']
// this creates the variable from the contents of $row['name']
$$row['name'] = $row['content'];
}
echo $username; // should return 'a'
简而言之,行的"name"列包含变量名,"content"列包含值。
选项表是关系数据库系统中使用的一个常见抽象,称为实体属性值。这是一篇可爱的文章:
http://mikesmithers.wordpress.com/2013/12/22/the-anti-pattern-eavil-database-design
以下是如何查询options
表:
<?php
$query = mysqli_query($con, "SELECT * FROM options");
?>
以下是如何将信息捕获到PHP数组中:
<?php
$user = array();
while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
$user[ $row['name'] ] = $row['content'];
}
var_dump( $user );
?>
以下是您访问数据的方式:
<html>
<p>username: <?= $user['username'] ?></p>
<p>password: <?= $user['password'] ?></p>
</html>
请注意,您可能需要在选项表中添加一个user_id
列。这就是EAV表能够支持多个用户的方式。