FOUND_ROWS()在mySQL中的错误结果

Question

我一共有6排。当我执行一个查询(例如SELECT * from表)时，我有

retrieved

• LIMIT 3 => FOUND_ROWS()给出3 => 3行为retrieved

• LIMIT 1，3 => FOUND_ROWS()给出4 => 3行为retrieved

• LIMIT 2，3 => FOUND_ROWS()给出5 => 3行为retrieved

• LIMIT 3，3 => => 3行为retrieved

• LIMIT 4，3 => ()给出6#en22 20#2行被检索

知道是什么原因导致了这种奇怪的行为吗？

SQL查询

SELECT `places`.*, `category`.*, COUNT(places_reviews.place_id) AS num_reviews, (places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating, FOUND_ROWS() AS num_rows FROM (`places`) JOIN `category` ON `places`.`category_id` = `category`.`category_id` LEFT JOIN `places_reviews` ON `places_reviews`.`place_id` = `places`.`id` LEFT JOIN `places_popularity` ON `places_popularity`.`place_id` = `places`.`id` WHERE `places`.`category_id` = 1 AND `places`.`name` LIKE '%%' GROUP BY `places`.`id` ORDER BY `id` desc LIMIT 3

或块中的：

SELECT `places`.*, `category`.*, 
COUNT(places_reviews.place_id) AS num_reviews, 
(places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating, FOUND_ROWS() AS num_rows FROM (`places`) 
JOIN `category` ON `places`.`category_id` = `category`.`category_id` 
LEFT JOIN `places_reviews` ON `places_reviews`.`place_id` = `places`.`id` 
LEFT JOIN `places_popularity` ON `places_popularity`.`place_id` = `places`.`id` 
WHERE `places`.`category_id` = 1 
    AND `places`.`name` LIKE '%%' 
GROUP BY `places`.`id` 
ORDER BY `id` desc LIMIT 3

Answer 1

编辑：

这就是你要找的东西：http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_found-rows

所以在您的查询中：

选择sql_calc_found_rows .

Answer 2

试着这样做

SELECT sql_calc_found_rows `places`.*, `category`.*, 
COUNT(places_reviews.place_id) AS num_reviews, 
(places_popularity.rating_1 + 2*places_popularity.rating_2 + 3*places_popularity.rating_3 + 4*places_popularity.rating_4 + 5*places_popularity.rating_5)/(places_popularity.rating_1 + places_popularity.rating_2 + places_popularity.rating_3 + places_popularity.rating_4 + places_popularity.rating_5) AS average_rating FROM (`places`) 
JOIN `category` ON `places`.`category_id` = `category`.`category_id` 
LEFT JOIN `places_reviews` ON `places_reviews`.`place_id` = `places`.`id` 
LEFT JOIN `places_popularity` ON `places_popularity`.`place_id` = `places`.`id` 
WHERE `places`.`category_id` = 1 
    AND `places`.`name` LIKE '%%' 
GROUP BY `places`.`id` 
ORDER BY `id` desc LIMIT 3;

select found_rows();

Answer 3

最快的解决方案是按以下方式对实际查询进行子查询：

SELECT SQL_CALC_FOUND_ROWS * FROM (SELECT whatever FROM whatever WHERE whatever LIMIT whatever) ax; 
select FOUND_ROWS();

现在你将得到正确的结果。主要原因是SQL_CALC_FOUND_ROWS主要跟踪已找到的行(即没有LIMITS)，没有返回行。