MySQL查詢組按性別和類型付款

Question

我有一個表訪問。 我想說明的類型的薪酬和性別的訪問

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 ; 

我想有這樣的結果：

 
| 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stucture

 

+-----------------+-------------+------+-----+---------+-------+ 
| 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

沒有試圖真的很難懂列名：

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