从多个递增序列中选择最大值的总和

对于这组数据：

time_stamp count 
1467820429 6 * 
1467820428 5 
1467820427 4 
1467820426 3 
1467820416 2 
1467820415 1 
1467820413 0 
1467820412 3 * 
1467820411 2 
1467820409 1 
1467820408 0 
1467820405 1 * 
1467820404 0 
1467820400 5 *

答案= 6 + 3 + 1 + 5 = 15

我怎么能写一个MySQL兼容SQL语句来acheve这个

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2016-07-06 Kevin

你能解释逻辑，你怎么确定他们为最大值 –

这是要痛'mysql的做'，当其他主要的'RDBMS'具有一些很好的功能来以更简单的方式做这样的事情时 –

下一个值比它低 – Kevin

SELECT SUM(a.cnt) 
    FROM 
    (SELECT x.* 
      , MIN(y.time_stamp) next 
     FROM my_table x 
     LEFT 
     JOIN my_table y 
      ON y.time_stamp > x.time_stamp 
     GROUP 
      BY x.time_stamp 
    ) a 
    LEFT 
    JOIN my_table b 
    ON b.time_stamp = a.next 
    AND b.cnt > a.cnt 
WHERE b.cnt IS NULL;

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2016-07-06 16:29:53 Strawberry

-1

您需要GROUP BY和HAVING，如下所示：

select sum (count) 
from table 
group by time_stamp 
having count = max(count)

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2016-07-06 16:13:48

** NMDV **，这不会得到预期的结果 –

你可以用下面的方法得到它

mysql> select time_stamp,count,if (count=0,@curRank :=0,@curRank := @curRank + 1) as rank from ff,(SELECT @curRank := 0) r; 
+------------+-------+------+ 
| time_stamp | count | rank | 
+------------+-------+------+ 
| 1467820429 |  6 | 1 | 
| 1467820428 |  5 | 2 | 
| 1467820427 |  4 | 3 | 
| 1467820426 |  3 | 4 | 
| 1467820415 |  2 | 5 | 
| 1467820415 |  1 | 6 | 
| 1467820413 |  0 | 0 | 
| 1467820412 |  3 | 1 | 
| 1467820411 |  2 | 2 | 
| 1467820409 |  1 | 3 | 
| 1467820408 |  0 | 0 | 
| 1467820405 |  1 | 1 | 
| 1467820404 |  0 | 0 | 
| 1467820408 |  5 | 1 | 
+------------+-------+------+ 
14 rows in set (0.00 sec) 

mysql> SELECT * FROM (select time_stamp,count,if (count=0,@curRank :=0,@curRank := @curRank + 1) as rank from ff,(SELECT @curRank := 0) r) t WHERE rank=1; 
+------------+-------+------+ 
| time_stamp | count | rank | 
+------------+-------+------+ 
| 1467820408 |  5 | 1 | 
| 1467820412 |  3 | 1 | 
| 1467820429 |  6 | 1 | 
| 1467820405 |  1 | 1 | 
+------------+-------+------+ 
4 rows in set (0.00 sec) 

    mysql> SELECT sum(count) as total FROM 
(select time_stamp,count,if (count=0,@curRank :=0,@curRank := @curRank + 1) as rank from ff, 
(SELECT @curRank := 0) r) t WHERE rank=1; 
    +-------+ 
    | total | 
    +-------+ 
    | 15 | 
    +-------+ 
    1 row in set (0.00 sec)

你可以用简单的内部查询

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2016-07-06 16:21:57

没有ORDER BY子句，这不能可靠地工作。我注意到你的数据集（目前）与OP的不同。 – Strawberry

@Strawberry是同意的。简单的我们可以使用时间戳命令。对此没有伤害 –

另外，测试不是'当count = 0'时（尽管这也是我的误解，并且在这种情况下会起作用）。根据评论，测试是当'下一个比以前小' – Strawberry

得到它，因为我在评论中提到，没有有效的方式Mysql做到这一点ATLEAST据我所知

试试这个

SELECT Sum(CASE 
      WHEN `count` >= prev_cnt THEN `count` 
      ELSE 0 
      END) 
FROM (SELECT *, 
       IFnull((SELECT `count` 
         FROM yourtable b 
         WHERE a.`time_stamp` < b.`time_stamp` 
         ORDER BY `time_stamp` LIMIT 1), `count`) AS prev_cnt 
     FROM yourtable a) c

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2016-07-06 16:24:03

谢谢@Prdp – Kevin

您需要确定值何时变化。一种获得上一个值的方法使用变量：

select sum(count) 
from (select t.*, 
      (if((@old_c := @c) is null, 0, -- never happens 
       if((@c := count) is not null, @old_c, @old_c) 
       ) 
      ) as prev_count 
     from t cross join 
      (select @c := -1) params 
     order by time_stamp 
    ) t 
where prev_count >= count;

获得前一个计数的表达式有点复杂。 MySQL不保证表达式求值的顺序，因此赋予新值count和返回旧值需要在一个表达式中。

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2016-07-06 16:25:17

有没有一个逗号？还有过多的't's？即使纠正这些错误，并忽略了括号，我认为这是不对的。 – Strawberry

@Strawberry。。。我确实认为这是正确的（我修复了一个缺失的结论），并且应该有最好的表现。 –

't。* （if（'can not be correct。I agree if if it can be done（I could not find out how）then using variables could be faster，but there are errors occurred here。 – Strawberry

-2

非常简单的解决方案： 1将列TIME-STAMP的滞后

2- take the difference of orif time_stamp column and the lag column 

3- sum the values of count after filtering out the records for -1 

    +------------+-------+------+-----------------+-------+ 
| a   | b  | c | d    | a-d | 
| time_stamp | count | flag | lag_time_stamp | diff | 
| 1467820429 | 6  | * | nulll   | null | 
| 1467820428 | 5  |  | 1467820429  | -1 | 
| 1467820427 | 4  |  | 1467820428  | -1 | 
| 1467820426 | 3  |  | 1467820427  | -1 | 
| 1467820416 | 2  | * | 1467820426  | -10 | 
| 1467820415 | 1  |  | 1467820416  | -1 | 
| 1467820413 | 3  | * | 1467820415  | -2 | 
| 1467820412 | 3  |  | 1467820413  | -1 | 
| 1467820411 | 2  |  | 1467820412  | -1 | 
| 1467820409 | 1  | * | 1467820411  | -2 | 
| 1467820408 | 0  |  | 1467820409  | -1 | 
| 1467820405 | 1  | * | 1467820408  | -3 | 
| 1467820404 | 0  |  | 1467820405  | -1 | 
| 1467820400 | 5  | * | 1467820404  | -4 | 
+------------+-------+------+-----------------+-------+ 

--sum the values of the table that we got after filtering the records for -1 

+------------+-------+ 
| time_stamp | count | 
+------------+-------+ 
| 1467820429 |  6 | 
| 1467820416 |  2 | 
| 1467820413 |  3 | 
| 1467820409 |  1 | 
| 1467820405 |  1 | 
| 1467820400 |  5 | 
+------------+-------+

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2016-07-06 17:59:17

我想你'重新混淆'简单'与'错误' – Strawberry

从多个递增序列中选择最大值的总和

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