SQL查询连续项目检查总数

Question

SQLITE/SQL 查询连续记录总数，PFA 图像。 在这里，我在表中有 1 和 0。 假设我想检查正好是 3 的倍数的 1 的总数，或者你可以说 countinuous。如果出现零，则不予考虑。

示例 A：-（水平显示的列值）

1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 

所以出现的总数是 8 任意顺序除以 3。

[编辑 2] 将第一列视为毫秒，并按顺序排列

Answer 1

考虑到我们可以使用deviceid 对行进行排序和对序列进行计数，因此尝试回答您的问题

请注意，为了工作，您至少应该拥有 sqlite 3.25 版本，因为这需要窗口函数

创建一个带有自动增量deviceid 列的测试表，这将模拟您自己的数据

CREATE TABLE "test" (
    "deviceid"  INTEGER,
    "password"  INTEGER,
    PRIMARY KEY("deviceid" AUTOINCREMENT)
);

按照您自己的顺序插入行1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1

DELETE FROM test;
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (0); 
INSERT INTO test (password) VALUES (1); 
INSERT INTO test (password) VALUES (1);

解决方案：

select password, sum(div_floor) AS Number_of_3_occurences
FROM (
    select password, count(*) as cnt_sequence, cast(round(count(*) / 3 ) as int) as div_floor
    from (select test.*,
                 (row_number() over (order by deviceid) -
                  row_number() over (partition by password order by deviceid)
                 ) as grp
          from test
         ) t
    group by grp, password
) ttt
where  cnt_sequence>=3
group by password 

结果

password  Number_of_3_occurences
0         2
1         8