这非常快，而且是100%随机的，即使你有间隙。

将SELECT Count(*)中可用的行数计算为rows FROM TABLE 选择10个不同的随机数a_1,a_2，…，a_10在0到x之间 SELECT * FROM TABLE LIMIT 1 offset a_i for i=1，…，10

我在Bill Karwin的《SQL反模式》一书中发现了这个破解方法。

SELECT
  * 
FROM
  table_with_600k_rows
WHERE
  RAND( ) 
ORDER BY
  id DESC 
LIMIT 30;

Id是主键，按Id排序， 解释table_with_600k_rows，发现该行不扫描整个表

SELECT column FROM table
ORDER BY RAND()
LIMIT 10

这不是有效的解决方案，但确实有效

以下内容应该是快速的，公正的，独立于id列的。但是，它不能保证返回的行数与请求的行数匹配。

SELECT *
FROM t
WHERE RAND() < (SELECT 10 / COUNT(*) FROM t)

解释:假设你想要100行中的10行，那么每一行都有1/10的概率被选中，这可以通过WHERE RAND() < 0.1来实现。这种方法不能保证有10行;但是如果查询运行了足够多的次数，那么每次执行的平均行数将在10左右，并且表中的每一行都将被均匀地选择。

我是这样做的:

select * 
from table_with_600k_rows
where rand() < 10/600000
limit 10

我喜欢它，因为它不需要其他表，写起来很简单，执行起来非常快。

我需要一个查询从一个相当大的表中返回大量随机行。这是我想到的。首先获取最大记录id:

SELECT MAX(id) FROM table_name;

然后将该值代入:

SELECT * FROM table_name WHERE id > FLOOR(RAND() * max) LIMIT n;

Where max is the maximum record id in the table and n is the number of rows you want in your result set. The assumption is that there are no gaps in the record id's although I doubt it would affect the result if there were (haven't tried it though). I also created this stored procedure to be more generic; pass in the table name and number of rows to be returned. I'm running MySQL 5.5.38 on Windows 2008, 32GB, dual 3GHz E5450, and on a table with 17,361,264 rows it's fairly consistent at ~.03 sec / ~11 sec to return 1,000,000 rows. (times are from MySQL Workbench 6.1; you could also use CEIL instead of FLOOR in the 2nd select statement depending on your preference)

DELIMITER $$

USE [schema name] $$

DROP PROCEDURE IF EXISTS `random_rows` $$

CREATE PROCEDURE `random_rows`(IN tab_name VARCHAR(64), IN num_rows INT)
BEGIN

SET @t = CONCAT('SET @max=(SELECT MAX(id) FROM ',tab_name,')');
PREPARE stmt FROM @t;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

SET @t = CONCAT(
    'SELECT * FROM ',
    tab_name,
    ' WHERE id>FLOOR(RAND()*@max) LIMIT ',
    num_rows);

PREPARE stmt FROM @t;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
END
$$

then

CALL [schema name].random_rows([table name], n);