Question

我有三张关系表

1。表格权限（从用户名获取用户权限）

select branch_code,username from privilege where username='joko';

结果：

| branch_code  | username |
|--------------|----------|
| 'IDJK003MPV' | 'joko'   |
| 'IDJK001MAG' | 'joko'   |
| 'IDJK002MCL' | 'joko'   |
| 'IDBTNCTR'   | 'joko'   |

2。 table biodata_karyawan（此表有关具有branch_code的员工）

select branch_code,status_karyawan from biodata_karyawan;

结果：

| branch_code  | status_karyawan |
|--------------|-----------------|
| 'IDJB001BEC' | 'Aktif'         |
| 'IDJB001BEC' | 'Aktif'         |
| 'IDPA001PPA' | 'Aktif'         |
| 'IDJK001MAG' | 'Aktif'         |
| 'IDJB001BEC' | 'Aktif'         |
| 'IDJB001BEC' | 'Tidak Aktif'   |
| 'IDBTNCTR'   | 'Tidak Aktif'   |

第3。表格分支

select branch_code,branch_name from branch;

结果：

| branch_code  | branch_name              |
|--------------|--------------------------|
| 'IDJB001BEC' | 'BEC BANDUNG'            |
| 'IDJK001MAG' | 'MALL ARTHA GADING'      |
| 'IDJK001HO'  | 'HEAD OFFICE'            |
| 'IDPA001PPA' | 'MALL MATAHARI JAYAPURA' |
| 'IDBE001BGK' | 'BENCOOLEN INDAH MALL'   |
| 'IDJK002MPB' | 'METRO PASAR BARU'       |

我希望结果会像这些人一样:(仅记录样本）

| branch_code  | branch_name    | count |
|--------------|----------------|-------|
| 'IDJB001BEC' | 'MALL BANDUNG' | '5'   |
| 'IDJK001MLP' | 'MALL LIPPO'   | '2'   |
| 'IDJK002MPI' | 'MALL PURI'    | '0'   |
| 'IDJB002ZZZ' | 'MALL POSO'    | '0'   |

任何人都可以帮助我吗？谢谢你的时间。

Answer 1

在status_karyawan上使用count，然后按

分组

select bk.branch_code,b.branch_name,count(bk.status_karyawan) from
  biodata_karyawan bk join
  branch b on bk.branch_code = b.branch_code
  group by bk.branch_code,b.branch_name

这是sql fiddle

根据您的评论更新了查询

select p.branch_code,b.branch_name,count(bk.status_karyawan) from
 biodata_karyawan bk join
 branch b on bk.branch_code = b.branch_code
 join privilege p on b.branch_code = p.branch_code
 Where p.username = 'joko'
 group by p.branch_code,b.branch_name

更新fiddle

mysql左连接/右连接3表

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