arersine在sqlalchemy中的公式

Question

我的代码中有以下几行

    query = "SELECT id, " \
        "( 3959 * acos( cos( radians(37) ) * cos( radians( %(lat)i ) ) * " \
        "cos( radians( %(lng)i ) - radians(-122) ) + sin( radians(37) ) * " \
        "sin( radians( %(lat)i ) ) ) ) AS `distance` from message where" \
        " `distance` <= %(drange)d" % {'lat': float(lat), 'lng': float(lng), 'drange': int(drange)}
    print query
    messages = db.session.query(Message).from_statement(query).all()

我使用它时出现以下错误

OperationalError: (OperationalError) (1054, "Unknown column 'distance' in 'where clause'") 'SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( 0 ) ) * cos( radians( 0 ) - radians(-122) ) + sin( radians(37) ) * sin( radians( 0 ) ) ) ) AS `distance` from message where `distance` <= 50' ()

解决这个问题的正确方法是什么？

Answer 1

你不能distance在WHERE子句中引用命名的表达式()(我不知道这对于所有数据库系统是否通常都是正确的,但对于MySQL来说至少是这样).您可以HAVING改为使用(参见选项C).

选项:

A.在where子句中再次重复表达式:

SELECT id, (long_formula) as distance FROM message WHERE (long_formula) <= ...

B.使用嵌套查询:

SELECT * FROM 
(SELECT id, (long_formula) AS distance FROM message) inner_query 
WHERE distance <= ...

C.使用HAVING条款(我使用SQL多年,但并没有意识到HAVING,直到我读这个):

SELECT id, (long_formula) as distance FROM message HAVING distance <= ...