Php 使用两列将两个SQL表相互连接起来

问题：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

我试图将评分表（betyg_文章）中两列中的id“翻译”为另一个表（betyg_用户）中这些用户的真实姓名

评分表（betyg_论文）如下所示：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

用户表（betyg\u users）如下所示：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

PHP代码（到目前为止）：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

考官/主管栏（betyg_论文）与UID栏（betyg_用户）相对应

所需的输出应该是（使用Firstname/Lastname列）：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

考官：约翰·霍普金斯。主管：迈克·贝克

考官：约翰·霍普金斯。主管：迈克·贝克

非工作SQL查询：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

我没有尝试代码，您可能需要进行一些调整

更新：修复了id的错误

INNER JOIN betyg_users u2 ON e.Supervisor = u2.UID

更新2：

我已将+更改为CONCAT函数。显然，MySQL处理字符串的方式与SQL Server不同

我没有尝试代码，您可能需要进行一些调整

更新：修复了id的错误

INNER JOIN betyg_users u2 ON e.Supervisor = u2.UID

更新2：

我已将+更改为CONCAT函数。显然，MySQL处理字符串的方式与SQL Server不同。

试试这个

SELECT CONCAT(betyg_users.`Firstname`," ",betyg_users.`Lastname`) AS `examiner_name`,
CONCAT(betyg_users.`Firstname`," ",betyg_users.`Lastname`) AS `superviser_name`
FROM `betyg_essays`
INNER JOIN `betyg_users` ON betyg_essays.`Examiner` = betyg_users.`UID` 
INNER JOIN `betyg_users` ON betyg_essays.`Supervisor` = betyg_users.`UID`;

试试这个

SELECT CONCAT(betyg_users.`Firstname`," ",betyg_users.`Lastname`) AS `examiner_name`,
CONCAT(betyg_users.`Firstname`," ",betyg_users.`Lastname`) AS `superviser_name`
FROM `betyg_essays`
INNER JOIN `betyg_users` ON betyg_essays.`Examiner` = betyg_users.`UID` 
INNER JOIN `betyg_users` ON betyg_essays.`Supervisor` = betyg_users.`UID`;

---

在这里，您需要两次加入user_表，第一次是为了获得考官用户名，第二次是为了获得主管用户名

SELECT (g1.FirstName + ' ' + g1.LastName) AS Examiner,' , (g2.FirstName + ' ' + g2.LastName) AS Supervisor
FROM grading_table grade
INNER JOIN user_table g1 ON grade.examiner = g1.uid
INNER JOIN user_table g2 ON grade.supervisor = g2.uid

---

在这里，您需要两次加入user_表，第一次是为了获得考官用户名，第二次是为了获得主管用户名

SELECT (g1.FirstName + ' ' + g1.LastName) AS Examiner,' , (g2.FirstName + ' ' + g2.LastName) AS Supervisor
FROM grading_table grade
INNER JOIN user_table g1 ON grade.examiner = g1.uid
INNER JOIN user_table g2 ON grade.supervisor = g2.uid

---

最终解决方案：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

---

最终解决方案：

$query = 'SELECT * FROM betyg_essays JOIN betyg_users ON betyg_essays.Examiner = betyg_users.UID';

$query = "SELECT 
            (u1.Firstname + ' ' + u1.Lastname) AS Examiner, 
            (u2.Firstname + ' ' + u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

$query = "SELECT 
            CONCAT (u1.Firstname, ' ', u1.Lastname) AS Examiner, 
            CONCAT (u2.Firstname, ' ', u2.Lastname) AS Supervisor
        FROM betyg_essays grade
            INNER JOIN betyg_users u1 ON grade.Examiner = u1.UID
            INNER JOIN betyg_users u2 ON grade.Supervisor = u2.UID";

---

我已经尝试了你的代码，但是输出只给了我0和0。我已经更新了我在问题中使用的代码。@kexxcream我已经更改了我的答案，我不知道mysql不允许只对数字进行+操作：/我已经尝试了你的代码，但输出只给了我0和0。我已经更新了我在问题中使用的代码。@kexxcream我已经更改了我的答案，我不知道mysql不允许只对数字进行+操作：/谢谢，但我希望将考官和主管中的名字和姓氏结合起来。欢迎您。我已经更新了我的答案，它会给你名字和姓氏的组合谢谢，但我希望在考官和主管中结合名字和姓氏。欢迎你。我已经更新了我的答案，它会给你名字和姓氏的组合。这很好，但我希望在考官和主管中组合名字/姓氏。这很好，但我希望在考官和主管中组合名字/姓氏。