Question

我在复杂的查询中遇到此问题。这是简化版本：

SELECT sin(3.14) as s from my_table where s < 1

错误：＃1054 - 'where子句'
中的未知列''

Answer 1

HAVING会拯救你

SELECT *,(((acos(sin((".$lat."*pi()/180)) *sin((latitud*pi()/180))
+cos((".$lat."*pi()/180)) 
* cos((latitud*pi()/180)) 
* cos(((".$lng."- longitud)*pi()/180))))*180/pi())
*60*1.1515*1.609344) as distance 
FROM pois_data,pois_cat 
WHERE pois_data.idtipo=pois_cat.id AND latitud IS NOT NULL 

HAVING distance <1

Answer 2

SELECT sin(3.14) as s from my_table where sin(3.14) < 1;

Answer 3

在该表上创建一个VIEW，该表达式为列。

CREATE VIEW vw
SELECT sin(3.14) AS s FROM my_table;

SELECT * FROM vw WHERE s < 1;

或者在表格中创建一个虚拟列。

CREATE TABLE my_table (
    ...,
    s VIRTUAL FLOAT AS (sin(3.14))
);

MySQL中计算字段的未知列

3 个答案: