Question

我需要将以下两个工作查询转换为单个查询但所有内容我试着因为各种原因而死在我身上。我的最终结果是尝试列出所有现有软件，并显示已安装的软件以及未针对要查询的特定PC安装的软件。对于已安装的软件，请列出名称，否则显示名称为NULL。我在where子句中尝试了一些子选择语句，它给了我一个没有错误的结果，但没有给出正确的结果。任何帮助表示赞赏。

qry1

SELECT device_software.sw_id

FROM Software_device LEFT JOIN Device ON Software_device.d_id = Device.d_id

WHERE Device.d_id = 1;

qry2

SELECT Software.name, Software.sw_id, qry1.sw_id

FROM software LEFT JOIN qry1 ON software.sw_id = qry1.sw_id;

设备表

------------------
| name  | d_id  |
------------------
| PC1   | 1     |
| PC2   | 2     |
| PC3   | 3     |
------------------

软件表

------------------
| name  | sw_id |
------------------
| SW_a  | A     |
| SW_b  | B     |
| SW_c  | C     |
| SW_d  | D     |
------------------

Software_Device Table（多对多）

------------------
| d_id  | sw_id |
------------------
| 1     | A     |
| 1     | B     |
| 2     | A     |
| 2     | B     |
| 2     | C     |
------------------

结果我正在寻找...... （PC1上已安装和未安装的软件）

---------------------------------
| Sotfware  | pc_id |   name    |
---------------------------------
| SW_a      | 1     |   PC1     |
| SW_b      | 1     |   PC1     |
| SW_c      | NULL  |   NULL    |
| SW_d      | NULL  |   NULL    |
---------------------------------

我列出了mysql和sql标签，因为我认为这不重要，但是如果它确实存在，我正在使用mysql。

Answer 1

SELECT 
   s.name AS software, 
   IF((SELECT COUNT(sw.d_id) FROM software_device sw WHERE sw.sw_id = s.sw_id AND d_id = 1) > 0, 1, NULL) AS pc_id, 
    (SELECT d.name FROM device d INNER JOIN software_device sw ON d.d_id = sw.d_id WHERE sw.sw_id = s.sw_id AND d.d_id = 1)  AS name
FROM 
   software s
ORDER BY s.name

编辑2：也许不是最有效/最美丽的，但它有效

Answer 2

如果RomanKonz的答案肯定很接近，只需将“where”移动到“on”，让左连接正常工作。

select software.name as Software,
    device.d_id as pc_id,
    device.name as name
from software
    left join device_software
        on device_software.sw_id = software.sw_id
       and **device_software**.d_id = 1
    left join device
        on device_software.d_id = device.d_id
;

Answer 3

以下内容可以满足您的需求：

select s.`name` as `Software`, d.`d_id`, d.`name` as `Device` 
from software s
      left outer join software_device sd 
          on sd.`sw_id` = s.`sw_id` and sd.`d_id` = 1     
      left outer join device d using (`d_id`)  
order by s.`name`

使用WHERE选择LEFT / RIGHT JOIN并返回NULL

3 个答案: