MySQLの - 同じテーブル

Question

から複数の列によってグループが私はそう

に...次

sno | booking_id | room_type | gender | age | amount | days 
1 | 2016JUL001 | AC  | Male | 25 | 1000 | 15 
2 | 2016JUL001 | AC  | Male | 42 | 1000 | 15 
3 | 2016JUL001 | AC  | Male | 28 | 1000 | 15 
4 | 2016JUL010 | N AC  | Female | 45 | 1000 | 15 
5 | 2016JUL010 | N AC  | Female | 46 | 1000 | 15 
6 | 2016JUL005 | N AC  | Male | 28 | 1000 | 15 
7 | 2016JUL005 | N AC  | Female | 35 | 1000 | 15 
8 | 2016JUL009 | AC  | Female | 26 | 1000 | 15 
9 | 2016JUL009 | AC  | Female | 25 | 1000 | 15 

のような表を持って

の予想される出力[私は性別を取得したい場合=「女性」]

sno | booking_id | room_type | gender | age | amount | days 
4 | 2016JUL010 | N AC  | Female | 45 | 1000 | 15 
5 | 2016JUL010 | N AC  | Female | 46 | 1000 | 15 
8 | 2016JUL009 | AC  | Female | 26 | 1000 | 15 
9 | 2016JUL009 | AC  | Female | 25 | 1000 | 15 

の予想される出力

sno | booking_id | room_type | gender | age | amount | days 
1 | 2016JUL001 | AC  | Male | 25 | 1000 | 15 
2 | 2016JUL001 | AC  | Male | 42 | 1000 | 15 
3 | 2016JUL001 | AC  | Male | 28 | 1000 | 15 

[私は= 'マーレ性別を取得したい場合]

期待出力

sno | booking_id | room_type | gender | age | amount | days 
6 | 2016JUL005 | N AC  | Male | 28 | 1000 | 15 
7 | 2016JUL005 | N AC  | Female | 35 | 1000 | 15 

NOTE [私は性別=「男」AND性別=「女性」を取得したい場合]：私は3つの別々の個々のクエリは、上記の出力にで

感謝を取得したいです

Answer 1

だ、私は

クエリ[ '女性']

SELECT sno, bd.booking_id, bd.room_type, bd.gender, bd.age 
FROM customer_data bd 
INNER JOIN ( 
    SELECT booking_id, GROUP_CONCAT(DISTINCT gender) AS g 
    FROM customer_data 
    WHERE gender!='' AND age>0 
    GROUP BY booking_id 
    HAVING COUNT(booking_id) > 1 
    ORDER BY booking_id ASC, gender DESC 
) cbd 
WHERE cbd.booking_id = bd.booking_id AND cbd.g = 'Female' 

は、クエリ[必要なデータを取得することができています'男']

SELECT sno, bd.booking_id, bd.room_type, bd.gender, bd.age 
FROM customer_data bd 
INNER JOIN ( 
    SELECT booking_id, GROUP_CONCAT(DISTINCT gender) AS g 
    FROM customer_data 
    WHERE gender!='' AND age>0 
    GROUP BY booking_id 
    HAVING COUNT(booking_id) > 1 
    ORDER BY booking_id ASC, gender DESC 
) cbd 
WHERE cbd.booking_id = bd.booking_id AND cbd.g ='Male' 

クエリ[ '男と女']

SELECT sno, bd.booking_id, bd.room_type, bd.gender, bd.age 
FROM customer_data bd 
INNER JOIN ( 
    SELECT booking_id, GROUP_CONCAT(DISTINCT gender ORDER BY gender DESC) AS g 
    FROM customer_data 
    WHERE gender!='' AND age>0 
    GROUP BY booking_id 
    HAVING COUNT(booking_id) > 1 
    ORDER BY booking_id ASC, gender DESC 
) cbd 
WHERE cbd.booking_id = bd.booking_id AND cbd.g = 'Male,Female' 

Answer 2

まずクエリを進める：

SELECT sno, booking_id, room_type, gender, age 
FROM customer_data 
WHERE booking_id IN (SELECT booking_id FROM customer_data 
         WHERE gender='female' AND age>0 and RIGHT(booking_id,1) <> '1' 
         GROUP BY booking_id HAVING COUNT(*) > 1) 
ORDER BY booking_id ASC, age ASC 

第二：

SELECT sno, booking_id, room_type, gender, age 
FROM customer_data 
WHERE booking_id IN (SELECT booking_id FROM customer_data 
         WHERE gender='male' AND age>0 
         GROUP BY booking_id HAVING COUNT(*) > 1) 
ORDER BY booking_id ASC, age ASC 

そして第三に：

SELECT sno, booking_id, room_type, gender, age 
FROM customer_data 
WHERE booking_id IN (SELECT booking_id FROM customer_data 
         WHERE gender IN('male','female') AND age>0 
         GROUP BY booking_id HAVING COUNT(distinct gender) = 2) 
ORDER BY booking_id ASC, age ASC 

最初の二つであなただけの1性別を持っているだけbooking_idを望んでいた場合は、having句に追加します。

AND COUNT(distinct gender) = 1 

Answer 3

あなたのスキーマがやや不備表示されます。それにもかかわらず、ここにしようと多くの後に考えることを何か...

SELECT booking_id 
    , COUNT(DISTINCT gender) x 
    FROM customer_data 
WHERE gender IN ('Male','Female') <-- not strictly necessary if there are only two genders. 
GROUP 
    BY booking_id;