錯誤的結果與mySQL中的FOUND_ROWS（）

Question

我總共有6行。當我做一個查詢（比如選擇表*），我有

• LIMIT 3 => FOUND_ROWS（）給出3 => 3行檢索
• LIMIT 1，3 => FOUND_ROWS（），得到4- => 3行檢索
• LIMIT 2，3 => FOUND_ROWS（）給出5 => 3行檢索
• LIMIT 3,3 => FOUND_ROWS（）給出6 => 3行檢索
• LIMIT 4，3 => FOUND_ROWS（）給出6 => 2行被檢索

任何想法這種奇怪行爲的原因是什麼？

SQL查詢

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 

或者在一個塊：

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

編輯：

這是你在找什麼： http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_found-rows

因此，在您查詢：

SELECT SQL_CALC_FOUND_ROWS .....

Answer 2

嘗試以這種方式

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

最快的解決辦法是子查詢實際查詢是這樣的：

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

現在您將得到正確的結果。主要原因是SQL_CALC_FOUND_ROWS主要跟蹤找到的行（即沒有0​​）而不是返回的行。