mysql - Sql使用计数和位置显示所有数据

Question

我有 2 表银行和申请人，如何查询显示所有银行并计算使用该银行的申请人数量

Bank :

| id | Bank_desc    
|----|---------
| 1  | Ambank  
| 2  | Maybank  
| 3  | RHB BANK 
| 4  | OSBC     

[申请人][2] :

| id | Name | Bank|STEP|
|----|------|----|----|
|  1 | JACK |  3 | W1 |
|  2 | ANDY |  1 | W4 |
|  3 | VOID |  1 | W1 |
|  4 | RAY  |  1 | W5 |

我想显示所有银行列表以及申请人总数，但仅限于步骤 W1 或 W4 中的申请人，
[结果][2] :

| id | Bank     |Total|
|----|----------|----|
|  1 | Ambank   |  2 |
| 2  | Maybank  |  0 |
| 3  | RHB BANK |  1 |
| 4  | OSBC     |  0 |

我试过这个 sql

 SELECT
 b.id,
 b.Bank, COUNT(a.Bank) AS Total FROM Bank b LEFT JOIN Applicant a ON b.id = a.Bank WHERE a.step ='W1' or a.step='W4 GROUP BY b.id, b.Bank;

但结果显示只有银行有值(value)，而不是所有银行

[结果][2] :

| id | Bank     |Total|
|----|----------|----|
|  1 | Ambank   |  2 |
| 2  | RHB BANK |  1 |

如果我删除 WHERE a.step ='W1' 或 a.step='W4

它将显示所有银行列表，但也会显示所有步骤

任何帮助将不胜感激，在此先感谢!

DEMO SQL

Answer 1

你的问题是你的 WHERE条款不包括在这两个步骤中没有客户的任何银行。您需要更改查询以改用条件聚合:

WITH Bank AS (
    SELECT 1 AS id, 'Ambank' AS Bank UNION ALL
    SELECT 2, 'Maybank'  UNION ALL
    SELECT 3, 'RHB BANK' UNION ALL
    SELECT 4, 'OSBC'
),
Applicant AS (
    SELECT 1 AS id, 'JACK' AS Name, 3 AS Bank, 'W1' AS Step UNION ALL
    SELECT 2, 'ANDY', 1,'W1' UNION ALL
     SELECT 3, 'ROY', 1,'W4' UNION ALL
    SELECT 4, 'VOID', 1,'W5'
)

SELECT
    b.id,
    b.Bank,
    SUM(CASE WHEN a.step = 'W1' OR a.step = 'W4' THEN 1 ELSE 0 END) AS Total
FROM Bank b
LEFT JOIN Applicant a
    ON b.id = a.Bank
GROUP BY
    b.id,
    b.Bank
ORDER BY
    b.id;

输出:

id  Bank        Total
1   Ambank      2
2   Maybank     0
3   RHB BANK    1
4   OSBC        0

Updated DBFiddle