此 mySQL 查询存在问题：（使用带有 AS 子句的 WHERE）

Question

我的 SQL 查询工作正常，直到我尝试添加 'WHERE distance << 10' 和“计算块 AS 距离”分别位于第 4 行和第 10 行。知道我该如何修复它吗？谢谢！

Unknown column 'distance' in 'where clause'

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, 
6371 * acos( cos( radians(places.lat) ) * cos( radians( 1.29315 ) ) * cos( radians( 103.827164 ) - radians(places.lng) ) + sin( radians(places.lat) ) * sin( radians( 1.29315 ) ) ) AS distance 
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 `distance` < 5 AND places.name LIKE '%%' GROUP 
BY places.id 
ORDER BY id desc 
LIMIT 5

Answer 1

您需要将公式放入 WHERE 子句中，而不是使用别名距离。在 SQL 查询中，WHERE 子句在 SELECT 语句之前计算，因此别名（在本例中为 distance）尚不存在。您的 SQL 语句如下所示：

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, 
6371 * acos( cos( radians(places.lat) ) * cos( radians( 1.29315 ) ) * cos( radians( 103.827164 ) - radians(places.lng) ) + sin( radians(places.lat) ) * sin( radians( 1.29315 ) ) ) AS distance 
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 (6371 * acos( cos( radians(places.lat) ) * cos( radians( 1.29315 ) ) * cos( radians( 103.827164 ) - radians(places.lng) ) + sin( radians(places.lat) ) * sin( radians( 1.29315 ) ) )) < 5
   AND places.name LIKE '%%' GROUP 
BY places.id 
ORDER BY id desc 
LIMIT 5

通过名称引用 distance 的唯一方法是将您的语句包装起来，并将其放入新 SELECT 语句中的表中。例如：

SELECT *
FROM ( <insert your original query here without the WHERE distance= statement ) AS t
WHERE distance < 5