SQL開發實例和優化是怎樣的

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找出連續的數據如 1,2,3,48,50,51,52,53,67,68
找出連續的數字的起點和重點
1,3
48,48
51,53
67,68
create table test.range_problem(

a int not null,
primary key (a));
insert into test.range_problem values(1);

insert into test.range_problem values(2);

insert into test.range_problem values(3);

insert into test.range_problem values(48);

insert into test.range_problem values(50);

insert into test.range_problem values(51);

insert into test.range_problem values(52);

insert into test.range_problem values(53);

insert into test.range_problem values(66);

insert into test.range_problem values(67);

– 思路 1 找最后一個不連續的數
SELECT T1.A
FROM test.range_problem T1
WHERE NOT EXISTS
(
SELECT A
FROM test.range_problem T2
WHERE T2.A -1 = T1.A
)

– 思路 2 找所有行對應的結束值
SELECT ROW_NUMBER()OVER(ORDER BY TBASE.A) ID
,TBASE.A
,(SELECT Min(A)
FROM test.range_problem T1
WHERE NOT EXISTS
(
SELECT A
FROM test.range_problem T2
WHERE T2.A -1 = T1.A
)
AND T1.A = TBASE.A –T1.A 是 3,48,53,67 TBASE.A 是每行的值
) A_END

FROM test.range_problem TBASE

– 思路 3 分組后找出所有連續的起始值和結束值
SELECT MIN(A) A_START, A_END
FROM(
SELECT ROW_NUMBER()OVER(ORDER BY TBASE.A) ID
,TBASE.A
,(SELECT Min(A)
FROM test.range_problem T1
WHERE NOT EXISTS
(
SELECT A
FROM test.range_problem T2
WHERE T2.A -1 = T1.A
)
AND T1.A = TBASE.A –T1.A 是 3,48,53,67 TBASE.A 是每行的值
) A_END

FROM test.range_problem TBASE
) T
GROUP BY T.A_END

– 對上面語句的優化
WITH POT AS(
SELECT A
FROM test.range_problem T
WHERE NOT EXISTS
(SELECT A
FROM test.range_problem TC
WHERE TC.A -1 = T.A)
)

SELECT TBASE.A A_START,(SELECT MIN(A)
FROM POT T
WHERE TBASE.A = T.A
) A_END
FROM test.range_problem TBASE