Question

我总共有6行。当我进行查询（比如表格中的SELECT *）时，我有

LIMIT 3 =＆gt; FOUND_ROWS（）给出3 =＆gt;检索3行
LIMIT 1,3 =＆gt; FOUND_ROWS（）给出4 =＆gt;检索3行
LIMIT 2,3 =＆gt; FOUND_ROWS（）给出5 =＆gt;检索3行
LIMIT 3,3 =＆gt; FOUND_ROWS（）给出6 =＆gt;检索3行
LIMIT 4,3 =＆gt; FOUND_ROWS（）给出6 =＆gt;检索到2行

知道这种奇怪行为的原因是什么？

SQL QUERY

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

或在街区：

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：

SELECT sql_calc_found_rows .....

Answer 2

尝试这种方式

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

最快的解决方案是查询您的实际查询，如下所示：

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

现在您将获得正确的结果。主要原因是SQL_CALC_FOUND_ROWS主要跟踪找到的行（即没有LIMITS）而不是返回的行。

mySQL中FOUND_ROWS（）的结果错误

3 个答案: