Question

我在数据库中有2个表。

表1：用户 id，name，email，password，singleCourse，userType，

表2：userCourses id，fk_user_id，course_name

userType可以是1或2.

现在，如果userType = 1

继续从用户表中获取。

如果userType = 2

从userCourses表中获取包含多个记录的表。

需要查询逻辑方面的帮助。

除非有需要，否则查询不应从userCourses表中获取记录。

那么可以使用CASE吗？

以下是我尝试的内容：

$query = mysql_query("SELECT name.user,email.user,password.user,userType.user,
fk_user_id.course,course_name.course 
FROM  CASE userType.user WHEN 1 THEN FROM users.u 
WHEN 2 THEN users.user INNER JOIN userCourses.course 
ON (user.id = userCourses.fk_user_id)");

示例数据：

userCourses

1, John, john@test.com, 123456, 1, PHP
2, Nick, nick@test.com, 654321, 2, ''

userCourses：

1, 2, JAVASCRIPT
2, 2, C++
3, 2, Jquery

期望输出：

我只是想获取用户课程，但是如果usertype = 1

，使用所需的表，查询不应该转到另一个表来检查记录

Answer 1

我已发布如下的示例查询，根据您的要求进行修改

SELECT
(CASE field1
WHEN 'A' THEN 'value is A'
WHEN 'B' THEN 'value is B'
ELSE 'value is neither A or B'
END)
FROM your_table;

或

SELECT
(CASE
WHEN (field1 IS NULL) THEN 'value is NULL'
WHEN (field1 = 1) THEN 'value is 1'
ELSE 'value is neither NULL or 1'
END)
FROM your_table;

Answer 2

虽然由于未知原因您没有提供所需的输出，但看起来您只需要这个

SELECT u.id, u.name, u.email, u.password, u.userType, 
       COALESCE(NULLIF(u.singleCourse,''), c.course_name) course
  FROM users u LEFT JOIN userCourses c
    ON u.id = c.fk_user_id

或

SELECT id, name, email, password, userType, singleCourse course
  FROM users
 WHERE usertype = 1
UNION ALL
SELECT u.id, name, email, password, userType, c.course_name
  FROM users u JOIN userCourses c
    ON u.id = c.fk_user_id
 WHERE usertype = 2

示例输出：

+------+------+---------------+----------+----------+------------+
| id   | name | email         | password | userType | course     |
+------+------+---------------+----------+----------+------------+
|    2 | Nick | nick@test.com |   654321 |        2 | JAVASCRIPT |
|    2 | Nick | nick@test.com |   654321 |        2 | C++        |
|    2 | Nick | nick@test.com |   654321 |        2 | Jquery     |
|    1 | John | john@test.com |   123456 |        1 | PHP        |
+------+------+---------------+----------+----------+------------+

这是 SQLFiddle 演示

Answer 3

更新的帖子/代码：

<?php
 $query = mysqli_query("SELECT userType from user") or die("err:". mysqli_error());  
 $x = 1; 
 while($row = mysqli_fetch_array($query))
 {
    $userType[$x] = $row['userType']; // fetch all the userType in your table and put in an array
    $x++;
 }

for($x = 1; $x <= count($userType); $x++)
 { 

      if($userType[$x]="1") // determines if the user is "1" 
      { 
         $query = mysqli_query("SELECT * FROM user") or die("Error" . mysqli_error());
      }
      else
      { 
            $query = mysqli_query("SELECT u.*, c.* FROM user u LEFT OUTER JOIN userCourse c ON u.id = c.fk_user_id") or die(mysqli_error());
      }

      while(mysqli_fetch_array($query))
      {
        //code here whatever data you want to show
      }
 }

希望这会有所帮助。对早期代码的道歉。

决定字段值并从不同的表中获取记录

3 个答案: