使用 SQL 查询打印素数

本文介绍了使用 SQL 查询打印素数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我是 StackOverflow 的新手，遇到了一个打印 2 到 1000 的素数的查询.如果这是最有效的编码方式，我使用了以下查询需要输入.

I am new to StackOverflow and have got stuck with a query to print prime numbers from 2 to 1000. I have used the below query need input if this is the most efficient way to code it.

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;

我知道这个问题已经被问过很多次了，如果有的话，我请求更好的解决方案.更多关于它如何与 MySQL/MS SQL/PostgreSQL 一起工作的需要输入.

I know this question has been asked multiple times and I am requesting better solution if any. More over need input on how this works with MySQL/MS SQL/PostgreSQL.

任何帮助都会让我更好地理解.

Any help will make my understanding better.

推荐答案

在 PostgreSQL 中打印最多 1000 个素数的最快查询可能是:

In PostgreSQL probably the most fastest query that prints prime numbers up to 1000 is:

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 ;

在我的电脑上只用了 16 毫秒.

It took only 16 ms on my computer.

• 注意:从 en.wikipedia/wiki/复制的素数列表Prime_number并粘贴到这个长字符串中

如果您更喜欢 SQL，那么这可行

If you prefer SQL, then this works

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.

It's two times slower - 31 ms.

Ans 是 Oracle 的等效版本:

Ans an equivalent version for Oracle:

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 ) ;