如果你想要一个随机记录(不管id之间是否有空隙):

PREPARE stmt FROM 'SELECT * FROM `table_name` LIMIT 1 OFFSET ?';
SET @count = (SELECT
        FLOOR(RAND() * COUNT(*))
    FROM `table_name`);

EXECUTE stmt USING @count;

来源:https://www.warpconduit.net/2011/03/23/selecting-a-random-record-using-mysql-benchmark-results/评论- 1266

另一个简单的解决方案是对行进行排名，并随机获取其中之一，有了这个解决方案，你将不需要在表中有任何基于“Id”的列。

SELECT d.* FROM (
SELECT  t.*,  @rownum := @rownum + 1 AS rank
FROM mytable AS t,
    (SELECT @rownum := 0) AS r,
    (SELECT @cnt := (SELECT RAND() * (SELECT COUNT(*) FROM mytable))) AS n
) d WHERE rank >= @cnt LIMIT 10;

您可以根据需要更改限制值，以便访问尽可能多的行，但大多数情况下是连续的值。

然而，如果你不想要连续的随机值，那么你可以获取一个更大的样本并从中随机选择。就像……

SELECT * FROM (
SELECT d.* FROM (
    SELECT  c.*,  @rownum := @rownum + 1 AS rank
    FROM buildbrain.`commits` AS c,
        (SELECT @rownum := 0) AS r,
        (SELECT @cnt := (SELECT RAND() * (SELECT COUNT(*) FROM buildbrain.`commits`))) AS rnd
) d 
WHERE rank >= @cnt LIMIT 10000 
) t ORDER BY RAND() LIMIT 10;

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);

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

我想这是最好的办法了。

SELECT id, id * RAND( ) AS random_no, first_name, last_name
FROM user
ORDER BY random_no