CODEWITHSUNDEEP
phpmysqlsql
Nyxynyx
Nyxynyx

Reputation: 63599

Problem with this mySQL query: (using WHERE with AS clause)

My SQL query is working fine, until I try to add a 'WHERE distance < 10' and 'chunk-of-calculation AS distance' on 4th and 10th line respectively. Any idea how I can fix it? Thanks!

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

Upvotes: 2

Views: 565

Answers (1)

IAmTimCorey
IAmTimCorey

Reputation: 16757

You will need to put the formula in your WHERE clause instead of using the alias distance. In a SQL query, the WHERE clause is evaluated before the SELECT statement so the alias (in this case distance) does not exist yet. Here is what your SQL statement will look like:

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

The only way you could refer to distance by name would be to wrap your statement and make it into a table in a new SELECT statement. For example:

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

Upvotes: 3

Related Questions