我读过你不能在 WHERE 子句中使用 ALIAS,但我仍然没有一个好的替代解决方案来实现我在下面尝试做的事情。我可以用我的 DISTANCE 计算做什么,以便它在 WHERE 子句中可用?
SELECT n.nid AS nid, location.name AS location_name, (6371.0 * ACOS(SIN((location.latitude * RADIANS(1))) * SIN((28.755925 * RADIANS(1))) + COS((location.latitude * RADIANS(1))) * COS((28.755925 * RADIANS(1))) * COS((location.longitude * RADIANS(1)) - (-81.346395 * RADIANS(1))))) AS distance
FROM
node n
LEFT JOIN location_instance ON n.vid = location_instance.vid
LEFT JOIN location ON location_instance.lid = location.lid
WHERE (( (n.status = '1') AND (n.type IN ('locations')) AND (distance <= 100) ))
ORDER BY distance
LIMIT 10
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MySQL 扩展了HAVING子句的使用,可用于此目的。如果查询不是聚合查询,则HAVING仍会进行过滤——但它允许使用别名。
所以,你可以写:
SELECT n.nid AS nid, l.name AS l, (6371.0 * ACOS(SIN((l.latitude * RADIANS(1))) * SIN((28.755925 * RADIANS(1))) + COS((l.latitude * RADIANS(1))) * COS((28.755925 * RADIANS(1))) * COS((l.longitude * RADIANS(1)) - (-81.346395 * RADIANS(1))))) AS distance
FROM node n LEFT JOIn
location_instance li
ON n.vid = li.vid LEFT JOIN
location l
ON li.lid = l.lid
WHERE n.status = 1 AND n.type IN ('locations')
HAVING distance <= 100
ORDER BY distance
LIMIT 10;
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笔记:
status是一个数字,这似乎很可能,那么您应该与一个数字进行比较,而不是一个字符串。status并且type可以进入HAVING子句,但我怀疑它们在WHERE(我怀疑这HAVING可能会影响优化选择)中更好。