连接多个MySQL表的结果并使用PHP输出

使用PHP，我试图用MySQL数据库中两个不同表的数据填充HTML列表。各表的结构如下所示：

Table: "students"
+------------+------------+-----------+---------------+-------+
| student_id | first_name | last_name | city          | state |
+------------+------------+-----------+---------------+-------+
| 1          | Tobias     | Funke     | Newport Beach | CA    |
+------------+------------+-----------+---------------+-------+
| 2          | Bob        | Loblaw    | Laguna Beach  | CA    |
+------------+------------+-----------+---------------+-------+
| 3          | Ann        | Veal      | Bland         | CA    |
+------------+------------+-----------+---------------+-------+


Table: "students_current"
+------------+------------+---------------+
| student_id | school_id  | current_class |
+------------+------------+---------------+
| 1          | umass      | Sr            |
+------------+------------+---------------+
| 2          | ucla       | Jr            |
+------------+------------+---------------+
| 3          | ucla       | Fr            |
+------------+------------+---------------+

<li>
    <span class="first_name">Bob</span>
    <span class="last_name">Loblaw</span>
    <span class="city">Laguna Beach</span>
    <span class="state">CA</span>
    <span class="current_class">Jr</span>
</li>

 <li>
    <span class="first_name">Ann</span>
    <span class="last_name">Veal</span>
    <span class="city">Bland</span>
    <span class="state">CA</span>
    <span class="current_class">Fr</span>
</li>

我只想用与特定

学校id

匹配的记录填充列表

例如，如果我希望列表仅包含

学校id

为“ucla”的学生，则生成的HTML如下所示：

Table: "students"
+------------+------------+-----------+---------------+-------+
| student_id | first_name | last_name | city          | state |
+------------+------------+-----------+---------------+-------+
| 1          | Tobias     | Funke     | Newport Beach | CA    |
+------------+------------+-----------+---------------+-------+
| 2          | Bob        | Loblaw    | Laguna Beach  | CA    |
+------------+------------+-----------+---------------+-------+
| 3          | Ann        | Veal      | Bland         | CA    |
+------------+------------+-----------+---------------+-------+


Table: "students_current"
+------------+------------+---------------+
| student_id | school_id  | current_class |
+------------+------------+---------------+
| 1          | umass      | Sr            |
+------------+------------+---------------+
| 2          | ucla       | Jr            |
+------------+------------+---------------+
| 3          | ucla       | Fr            |
+------------+------------+---------------+

<li>
    <span class="first_name">Bob</span>
    <span class="last_name">Loblaw</span>
    <span class="city">Laguna Beach</span>
    <span class="state">CA</span>
    <span class="current_class">Jr</span>
</li>

 <li>
    <span class="first_name">Ann</span>
    <span class="last_name">Veal</span>
    <span class="city">Bland</span>
    <span class="state">CA</span>
    <span class="current_class">Fr</span>
</li>

上下快速移动
罗布劳
拉古纳海滩
加利福尼亚州
年少者


安
小牛肉
平淡的
加利福尼亚州
Fr

---

每个

项都将绑定到数据库中的特定

学生id

值。如何编写PHP，从数据库中选择/连接适当的记录？

使用

左连接：

SELECT *
FROM `students` s
    LEFT JOIN `students_current` sc ON s.`student_id` = sc.`student_id`
WHERE `school_id` = 'ucla'

使用
左连接
：

SELECT *
FROM `students` s
    LEFT JOIN `students_current` sc ON s.`student_id` = sc.`student_id`
WHERE `school_id` = 'ucla'

+1在SO问题示例数据中包含Bob Loblaw+1在SO问题示例数据中包含Bob Loblaw，我对不同类型的连接以及何时使用它们非常不熟悉。我一直在尝试使用一个内部连接，甚至没有正确地构造它。啊，我非常不熟悉不同类型的连接以及何时使用它们。我一直试图使用一个内部连接，甚至没有正确地构造它。