MySQL选择行直到达到固定的条件数

Question

我有这张桌子

id  fruit
---------
1   apple
2   banana <--
3   apple
4   apple
5   apple
6   apple
7   banana <----

8   apple
9   banana
10  apple

我想选择行直到找到2个香蕉，比如

SELECT id FROM table_fruit UNTIL number_of_bananas = 2

结果将是1,2,3,4,5,6,7

我怎么能实现这个目标？

感谢

我希望我可以给予回答我问题的所有人。我测试了所有这些，它们都完美地工作（得到了预期的结果）。

虽然Devart和ypercube的答案看起来有点复杂，我很难理解。

由于AnandPhadke是第一个提供有效解决方案的人，我会选择接受他的答案。

你们真棒，谢谢！

Answer 1

尝试此查询 -

SELECT id, fruit FROM (
  SELECT
    b.*, @b:=IF(b.fruit = 'banana', 1, 0) + @b AS banana_number
  FROM
    bananas b,
  (SELECT @b := 0) t
  ORDER BY id) t2
WHERE
  banana_number < 2 OR banana_number = 2 AND fruit = 'banana'

SQLFiddle demo

Answer 2

select * from tables where id <=

(
select id from (
select id from tables where fruit='banana'
order by id limit 2) a order by id desc limit 1
)

SQLFIDDLE DEMO

Answer 3

@Devart's answer是完美的，但它是我们可以使用的另一种选择：

SELECT * FROM table_fruit WHERE id <=
(
    SELECT id FROM 
    (SELECT id FROM table_fruit WHERE fruit='banana' ORDER BY id LIMIT 2) a
    ORDER BY ID DESC LIMIT 1
);

或使用MAX

SELECT * FROM table_fruit WHERE id <=
(
    SELECT MAX(id) FROM 
    (SELECT id FROM table_fruit WHERE fruit='banana' ORDER BY id LIMIT 2) a
);

See this SQLFiddle

Answer 4

select * from table_fruit where id <=
(
   select max(id) from
   (select id from table_fruit where fruit='banana' order by id limit 2) t
)

Answer 5

如果'banana'行少于2行，则会返回表格的所有行：

SELECT t.*
FROM table_fruit AS t
  JOIN
    ( SELECT MAX(id) AS id
      FROM
        ( SELECT id
          FROM table_fruit
          WHERE fruit = 'banana'
          ORDER BY id
            LIMIT 1 OFFSET 1
        ) AS lim2
    ) AS lim
    ON t.id <= lim.id
    OR lim.id IS NULL  ;