使用SQL查询打印Prime数字

时间：2016-05-15 05:28:58

标签： mysql sql-server postgresql select subquery

我是StackOverflow的新手，并且遇到了一个查询，打印从2到1000的素数。 如果这是最有效的编码方式，我已经使用了下面的查询需求输入。

WITH NUM AS (
    SELECT LEVEL N 
    FROM DUAL CONNECT BY LEVEL <= 1000
) 
SELECT LISTAGG(B.N,'-') WITHIN GROUP(ORDER BY B.N) AS PRIMES 
FROM (
    SELECT  N,
            CASE WHEN EXISTS (
                                SELECT NULL 
                                FROM NUM N_INNER 
                                WHERE N_INNER .N > 1 
                                AND N_INNER.N < NUM.N 
                                AND MOD(NUM.N, N_INNER.N)=0
                            ) THEN 
                'NO PRIME' 
            ELSE 
                'PRIME' 
            END IS_PRIME 
        FROM NUM
    ) B 
WHERE B.IS_PRIME='PRIME' 
AND B.N!=1;

我知道这个问题已被多次询问，我要求更好的解决方案。更多关于如何使用MySQL / MS SQL / PostgreSQL的输入。

任何帮助都会让我的理解变得更好。

20 个答案:

答案 0 :(得分：4)

在PostgreSQL中，打印质数高达1000的最快查询可能是：

SELECT regexp_split_to_table('2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997',E',')::int
AS x
;

我的电脑只用了16毫秒。

如果您更喜欢SQL，则可以使用

WITH x AS (
  SELECT * FROM generate_series( 2, 1000 ) x
)
SELECT x.x
FROM x
WHERE NOT EXISTS (
  SELECT 1 FROM x y
  WHERE x.x > y.x AND x.x % y.x = 0
)
;

它慢了两倍--31毫秒。

Ans for Oracle的等效版本：

WITH x AS(
    SELECT level+1 x
    FROM dual
    CONNECT BY LEVEL <= 999
)
SELECT x.x
FROM x
WHERE NOT EXISTS (
  SELECT 1 FROM x y
  WHERE x.x > y.x AND remainder( x.x, y.x) = 0
)
;

答案 1 :(得分：3)

最显着的改进是，不是从1到n检查，而是从1到n的平方根检查。

第二个主要优化是使用临时表来存储结果并首先检查它们。这样你就可以从1递增迭代到n，并且只检查n的1到平方根的已知素数（递归地执行，直到你有一个列表）。如果您按照这种方式进行操作，您可能希望在函数中设置质数检测，然后对数字序列生成器执行相同操作。

第二个虽然意味着扩展SQL，所以我不知道这是否符合您的要求。

对于postgresql，我会使用generate_series来生成数字列表。然后我会创建函数，然后将素数列表存储在临时表中，或者将它们以有序数组的形式传入和传出，然后将它们连接起来

答案 2 :(得分：1)

MariaDB（带序列插件）

类似于kordirkos算法：

select 2 as p union all
select n.seq
from seq_3_to_1000_step_2 n
where not exists (
    select 1
    from seq_3_to_32_step_2 q
    where q.seq < n.seq
      and n.seq mod q.seq = 0
);

使用LEFT JOIN：

select 2 as p union all
select n.seq
from seq_3_to_1000_step_2 n
left join seq_3_to_32_step_2 q
      on  q.seq < n.seq
      and n.seq mod q.seq = 0
where q.seq is null;

<强>的MySQL

MySQL中没有生成帮助程序的序列。因此必须首先创建序列表：

drop temporary table if exists n;
create temporary table if not exists n engine=memory
    select t2.c*100 + t1.c*10 + t0.c + 1 as seq from 
    (select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t0,
    (select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t1,
    (select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t2
    having seq > 2 and seq % 2 != 0;

drop temporary table if exists q;
create temporary table if not exists q engine=memory
    select *
    from n
    where seq <= 32;
alter table q add primary key seq (seq);

现在可以使用类似的查询：

select 2 as p union all
select n.seq
from n
where not exists (
    select 1
    from q
    where q.seq < n.seq
      and n.seq mod q.seq = 0
);

select 2 as p union all
select n.seq
from n
left join q
    on  q.seq < n.seq
    and n.seq mod q.seq = 0
where q.seq is null;

sqlfiddle

答案 3 :(得分：1)

在sqlite3上测试

WITH nums(n) AS 
(
    SELECT 1
    UNION ALL
    SELECT n + 1 FROM nums WHERE n < 100
)

SELECT n 
FROM (
  SELECT n FROM nums
) 
WHERE n NOT IN (
  SELECT n
  FROM nums 
  JOIN ( SELECT n AS n2 FROM nums )
  WHERE n  <> 1 
    AND n2 <> 1 
    AND n  <> n2 
    AND n2 <  n 
    AND n % n2 = 0
  ORDER BY n
)
AND n <> 1

在Vertica 8上测试

WITH seq AS (
  SELECT ROW_NUMBER() OVER() AS n 
  FROM (
      SELECT 1 
      FROM (
          SELECT date(0) + INTERVAL '1 second' AS i 
          UNION ALL
          SELECT date(0) + INTERVAL '100 seconds' AS i 
      ) _
      TIMESERIES tm AS '1 second' OVER(ORDER BY i)
  ) _
)
SELECT n 
FROM (SELECT n FROM seq) _  
WHERE n NOT IN (
  SELECT n FROM (
    SELECT s1.n AS n, s2.n AS n2
    FROM seq AS s1 
    CROSS JOIN seq AS s2
    ORDER BY n, n2
  ) _
  WHERE n  <> 1 
    AND n2 <> 1 
    AND n  <> n2 
    AND n2 <  n 
    AND n % n2 = 0
)
AND n <> 1
ORDER BY n

答案 4 :(得分：1)

Oracle，没有内在选择：

 with tmp(id)
as (
    select level  id from dual connect by level <= 100 
) select t1.id from tmp t1
 JOIN tmp t2
 on MOD(t1.id, t2.id) = 0
 group by t1.ID
 having count(t1.id) = 2
 order by t1.ID
 ;

答案 5 :(得分：1)

SELECT LISTAGG(PRIME_NUMBER,'&') WITHIN GROUP (ORDER BY PRIME_NUMBER) 
FROM
(
    SELECT L PRIME_NUMBER FROM
    (
        SELECT LEVEL L FROM DUAL CONNECT BY LEVEL <= 1000 ), 
        (
            SELECT LEVEL M FROM DUAL CONNECT BY LEVEL <= 1000
        ) WHERE M <= L 
        GROUP BY L 
        HAVING COUNT(CASE WHEN L/M = TRUNC(L/M) THEN 'Y' END
    ) = 2 
    ORDER BY L
);

答案 6 :(得分：1)

/* Below is my solution */

/* Step 1: Get all the numbers till 1000 */
with tempa as
(
  select level as Num
  from dual
  connect by level<=1000
),

/* Step 2: Get the Numbers for finding out the factors */

tempb as
(
    select a.NUm,b.Num as Num_1
    from tempa a , tempa b
    where b.Num<=a.Num
),

/*Step 3:If a number has exactly 2 factors, then it is a prime number */

tempc as
(
    select Num, sum(case when mod(num,num_1)=0 then 1 end) as Factor_COunt
    from tempb
    group by Num
)
select listagg(Num,'&') within group (order by Num)
from tempc
where Factor_COunt=2
;

答案 7 :(得分：0)

SELECT  GROUP_CONCAT(distinct PRIME_NUMBER SEPARATOR '&')
FROM (SELECT @prime:=@prime + 1 AS PRIME_NUMBER
    FROM information_schema.tables
    CROSS JOIN (SELECT @prime:=1) r
    WHERE @num <1000
    ) p
WHERE NOT EXISTS (
SELECT * FROM
    (SELECT @divisor := @divisor + 1 AS DIVISOR FROM
    information_schema.tables 
    CROSS JOIN (SELECT @divisor:=1) r1
     WHERE @divisor <=1000
    ) d
WHERE MOD(PRIME_NUMBER, DIVISOR) = 0 AND PRIME_NUMBER != DIVISOR) ;
    enter code here

说明：

  1. 两个内部 SELECT（SELECT @prime 和 SELECT @divisor）创建了两个列表。它们都包含从 1 到 1000 的数字。第一个列表是“潜在素数列表”，第二个是“除数列表”

  2. 然后，我们遍历潜在素数列表（外层 SELECT），对于这个列表中的每个数字，我们寻找除数（SELECT * FROM 子句），它可以在没有提示的情况下对数字进行除法，并且是不等于数字（WHERE MOD... 子句）。如果至少存在一个这样的除数，则该数不是质数且不会被选中（WHERE NOT EXISTS... 子句）。

答案 8 :(得分：0)

曾在甲骨文工作：

SELECT LISTAGG(a,'&')
     WITHIN GROUP (ORDER BY a) 
FROM(WITH x AS(
    SELECT level+1 x
    FROM dual
    CONNECT BY LEVEL <= 999
)

SELECT x.x as a
FROM x
WHERE NOT EXISTS (
  SELECT 1 FROM x y
  WHERE x.x > y.x AND remainder( x.x, y.x) = 0
));

答案 9 :(得分：0)

对于 MySQL 8 或更高版本

/* create a table with one row and that starts with  2 ends at 1000*/

SET cte_max_recursion_depth = 1001; /* works for MySQL 8.0*/
;WITH RECURSIVE sequence AS (
    SELECT 1 AS l
    UNION ALL
    SELECT l + 1 AS value
    FROM sequence
    WHERE sequence.l < 1000
),

/* create a caretesian product of a number to other numbers uptil this very number
so for example if there is a value 5 in a row then it creates these rows using the table below
(5,2), (5,3), (5,4), (5,5) */

J as (
SELECT (a.l) as m  , (b.l) as n 
FROM sequence a, sequence b
WHERE b.l <= a.l)
,

/*take a row from column 1 then divide it with other column values but group by column 1 first,
note the completely divisible count*/
f as 
( SELECT m , SUM(CASE WHEN mod(m,n) = 0 THEN 1  END) as fact
FROM  J 
GROUP BY m
HAVING fact = 2
ORDER BY m ASC /*this view return numbers in descending order so had to use order by*/
)

/* this is for string formatting, converting a column to a string with separator &*/
SELECT group_concat(m SEPARATOR '&') FROM f;

答案 10 :(得分：0)

这在 MySql 中对我有用：

select '2&3&5&7&11&13&17&19&23&29&31&37&41&43&47&53&59&61&67&71&73&79&83&89&97&101&103&107&109&113&127&131&137&139&149&151&157&163&167&173&179&181&191&193&197&199&211&223&227&229&233&239&241&251&257&263&269&271&277&281&283&293&307&311&313&317&331&337&347&349&353&359&367&373&379&383&389&397&401&409&419&421&431&433&439&443&449&457&461&463&467&479&487&491&499&503&509&521&523&541&547&557&563&569&571&577&587&593&599&601&607&613&617&619&631&641&643&647&653&659&661&673&677&683&691&701&709&719&727&733&739&743&751&757&761&769&773&787&797&809&811&821&823&827&829&839&853&857&859&863&877&881&883&887&907&911&919&929&937&941&947&953&967&971&977&983&991&997';

好吧，我知道上面的只是硬编码，您将能够运行问题，但这不是我们作为程序员应该追求的，所以这是我的 oracle 解决方案：

SELECT LISTAGG(L1,'&') WITHIN GROUP (ORDER BY L1) FROM (Select L1 FROM (SELECT LEVEL L1 FROM DUAL CONNECT BY LEVEL<=1000) Where L1 <> 1 MINUS select L1 from (SELECT LEVEL L1 FROM DUAL CONNECT BY LEVEL<=1000) A , (SELECT LEVEL L2 FROM DUAL CONNECT BY LEVEL<=1000) B Where L2<=L1 and MOD(L1,L2)=0 AND L1<>L2 AND L2<>1);

答案 11 :(得分：0)

-创建表prime_number_t 创建表prime_number_t（ integervalue_c整数，不为null主键 ）;

-将数据插入表prime_number_t 全部插入 变成prime_number_t（integervalue_c）个值（1） 成为prime_number_t（integervalue_c）个值（2） 成为prime_number_t（integervalue_c）个值（3） 成为prime_number_t（integervalue_c）个值（4） 变成prime_number_t（integervalue_c）个值（5） 变成prime_number_t（integervalue_c）个值（6） 变成prime_number_t（integervalue_c）个值（7） 变成prime_number_t（integervalue_c）个值（8） 成为prime_number_t（integervalue_c）个值（9） 成为prime_number_t（integervalue_c）个值（10）
从双重选择1；

COMMIT;

-编写SQL语句以确定以下哪个数字是质数 -相同的查询对REMAINDER函数也有效，而不是MOD函数 与cte_prime_number_t AS （ 选择integervalue_c 来自prime_number_t 按integervalue_c排序 ）， cte_maxval AS （ 选择max（integervalue_c）AS maxval FROM cte_prime_number_t ）， cte_level AS （ 选择LEVEL + 1作为等级 从双重， cte_maxval 按级别连接<= cte_maxval.maxval ） 选择DISTINCT cpnt.integervalue_c作为PrimeNumbers 从cte_prime_number_t cpnt lvl <=上的内部连接cte_level cl（从cte_maxval选择SELECT maxval） 不存在的地方（ 从cte_level cpn中选择1 其中cpnt.integervalue_c> cpn.lvl和mod（cpnt.integervalue_c，cpn.lvl）= 0 ） 按PrimeNumbers排序；

答案 12 :(得分：0)

这是在SQL Server中对我有用的。我试图减少嵌套循环的顺序。 

declare @var int
declare @i int
declare @result varchar (max)
set @var = 1
select @result = '2&3&5' --first few obvious prime numbers
while @var < 1000  --the first loop
begin
set @i = 3;
while @i <= @var/2  --the second loop which I attempted to reduce the order
begin
if @var%@i = 0
break;
if @i=@var/2 
begin
set @result = @result + '&' + CAST(@var AS VARCHAR)
break;
end
else 
set @i = @i + 1 
end
set @var = @var + 1;
end
print @result

答案 13 :(得分：0)

MySQL查询解决方案

我已经在下面的mysql中解决了这个问题：

SET @range = 1000;

SELECT GROUP_CONCAT(R2.n SEPARATOR '&')
FROM (
        SELECT @ctr2:=@ctr2+1 "n"
        FROM information_schema.tables R2IS1,
        information_schema.tables R2IS2,
        (SELECT @ctr2:=1) TI
        WHERE @ctr2<@range
     ) R2
WHERE NOT EXISTS (
                SELECT R1.n
                FROM (
                    SELECT @ctr1:=@ctr1+1 "n"
                    FROM information_schema.tables R1IS1,
                    information_schema.tables R1IS2,
                    (SELECT @ctr1:=1) I1
                    WHERE @ctr1<@range
                ) R1
                WHERE R2.n%R1.n=0 AND R2.n>R1.n
        )
  

注意：应增加 information_schema.tables 的数量，以扩大范围，例如如果范围是100000，则通过检查自己来设置信息表。

答案 14 :(得分：0)

SQL Server的最简单方法

DECLARE @range int = 1000, @x INT = 2, @y INT = 2 

While (@y <= @range)
BEGIN
 while (@x <= @y) 
 begin
    IF ((@y%@x) =0) 
    BEGIN
        IF (@x = @y) 
            PRINT @y
            break
    END
 IF ((@y%@x)<>0)   
 set @x = @x+1
 end  
set @x = 2
set @y = @y+1 
end

答案 15 :(得分：0)

一个简单的就可以这样

select level id1 from dual connect by level < 2001
minus
select distinct id1 from (select level id1 from dual connect by level < 46) t1 inner join (select level id2 from dual connect by level < 11) t2
on 1=1 where t1.id1> t2.id2 and mod(id1,id2)=0 and id2<>1

答案 16 :(得分：0)

For SQL Server We can use below CTE 

SET NOCOUNT ON

;WITH Prim AS
(
    SELECT 2 AS Value 
    UNION ALL
    SELECT t.Value+1 AS VAlue 
    FROM Prim t
    WHERE t.Value < 1000
)SELECT * 
FROM Prim t
WHERE NOT EXISTS(   SELECT 1 FROM prim t2
                WHERE t.Value % t2.Value = 0 
                AND t.Value != t2. Value)
OPTION (MAXRECURSION 0)

答案 17 :(得分：0)

PostgreSQL中的简单查询：

SELECT serA.el AS prime
FROM generate_series(2, 100) serA(el)
LEFT JOIN generate_series(2, 100) serB(el) ON serA.el >= POWER(serB.el, 2)
                                              AND serA.el % serB.el = 0
WHERE serB.el IS NULL

享受！ ：）

答案 18 :(得分：0)

以下代码可在SQL中查找素数

在本地服务器的SampleDB上测试

CREATE procedure sp_PrimeNumber(@number int)
as 
begin
declare @i int
declare @j int
declare @isPrime int
set @isPrime=1
set @i=2
set @j=2
while(@i<=@number)
begin
    while(@j<=@number)
    begin
        if((@i<>@j) and (@i%@j=0))
        begin
            set @isPrime=0
            break
        end
        else
        begin
            set @j=@j+1
        end
    end
    if(@isPrime=1)
    begin
        SELECT @i
    end
    set @isPrime=1
    set @i=@i+1
    set @j=2
end
end

我创建了一个存储过程，该存储过程具有一个@number参数，以查找直至给定数字的素数

为了获得素数，我们可以执行以下存储过程

EXECUTE sp_PrimeNumber 100  -- gives prime numbers up to 100

如果您不熟悉存储过程，并且想在SQL中查找素数，可以使用以下代码

在主数据库上测试

declare @i int
declare @j int
declare @isPrime int
set @isPrime=1
set @i=2
set @j=2
while(@i<=100)
begin
    while(@j<=100)
    begin
        if((@i<>@j) and (@i%@j=0))
        begin
            set @isPrime=0
            break
        end
        else
        begin
            set @j=@j+1
        end
    end
    if(@isPrime=1)
    begin
        SELECT @i
    end
    set @isPrime=1
    set @i=@i+1
    set @j=2
end

这段代码可以提供1到100之间的质数。如果我们想找到更多的质数，请在while循环中编辑@i和@j参数并执行

答案 19 :(得分：0)

MySQL代码：

DECLARE 
@i INT, 
@a INT, 
@count INT, 
@p nvarchar(max)
SET @i = 1 
WHILE (@i <= 1000) 
BEGIN SET @count = 0 
SET @a = 1 
WHILE (@a <= @i) 
BEGIN IF (@i % @a = 0) SET @count = @count + 1 SET @a = @a + 1 
END IF (@count = 2) SET @P = CONCAT(@P,CONCAT(@i,'&')) SET @i = @i + 1 
END
PRINT LEFT(@P, LEN(@P) - 1)
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