Question

我有一张桌子参观。我想按工资类型和性别显示访问次数

select jenis_bayar, (select count(id) 
from vw_kunjungan 
where waktu_kunjungan>='2015-02-12 00:00:00'AND waktu_kunjungan<='2015-02-12 23:59:00' 
and gender=1 
and puskesmas_id=1 
group by jenis_bayar) as total_lk,
         (select count(id) 
         from vw_kunjungan 
         where waktu_kunjungan>='2015-02-12 00:00:00'
         AND waktu_kunjungan<='2015-02-12 23:59:00' 
         and gender=0  
         and puskesmas_id=1 group by jenis_bayar) as total_pr,
        (select count(id) 
        from loket 
        where waktu_kunjungan>='2015-02-12 00:00:00'AND waktu_kunjungan<='2015-02-12 23:59:00' 
        and puskesmas_id=1 group by jenis_bayar) as total 
        from loket group by jenis_bayar ;

我希望得到这样的结果：

| type_of_pay |    man   |  woman   | total |
+-------------+----------+----------+-------+
| askes       |       12 |       10 |   22  |
| bayar       |       53 |       87 |   140 |
| bpjs_pbi    |       25 |       15 |   40  |
| gratis      |       10 |       10 |   20  |
| jamkesda    |       11 |       10 |   21  |
| kader       |       35 |       10 |   45  |
| non_pbi     |       60 |       50 |   110 |
+-------------+----------+----------+-------+

表结构


+-----------------+-------------+------+-----+---------+-------+
| Field           | Type        | Null | Key | Default | Extra |
+-----------------+-------------+------+-----+---------+-------+
| id              | int(11)     | NO   |     | 0       |       |
| loket_id        | int(11)     | NO   |     | 0       |       |
| no_registrasi   | varchar(45) | NO   |     | NULL    |       |
| nama            | varchar(25) | NO   |     | NULL    |       |
| nama_kk         | varchar(25) | YES  |     | NULL    |       |
| gender          | tinyint(1)  | NO   |     | NULL    |       |
| tgl_lahir       | date        | NO   |     | NULL    |       |
| waktu_kunjungan | datetime    | NO   |     | NULL    |       |
| jenis_kunjungan | tinyint(1)  | YES  |     | NULL    |       |
| puskesmas_id    | int(11)     | NO   |     | NULL    |       |
| jenis_bayar     | varchar(45) | NO   |     | NULL    |       |
| poli_id         | int(11)     | NO   |     | NULL    |       |
| _year           | bigint(21)  | YES  |     | NULL    |       |
| _month          | bigint(21)  | YES  |     | NULL    |       |
| _day            | bigint(17)  | YES  |     | NULL    |       |
| kelurahan_id    | int(11)     | NO   |     | NULL    |       |
+-----------------+-------------+------+-----+---------+-------+

Answer 1

没有真正难以理解您的列名：

select jenis_bayer  type_of_pay,
  sum(case when gender=1 then 1 else 0 end) man,
  sum(case when gender=1 then 0 else 1 end) woman,
  count(id) total
from vw_kunjungan
where waktu_kunjungan>='2015-02-12 00:00:00'AND waktu_kunjungan<='2015-02-12 23:59:00'
group by jenis_bayer

MySQL查询组按性别和类型付费

1 个答案: