我正在尝试在plsql中创建一个函数,以便在作为参数传递给函数的用户的1英里范围内找到5个用户。
这是我的代码:
CREATE OR REPLACE FUNCTION testfunction (integer) RETURNS integer as $$
DECLARE
myvar integer := $1;
mylon double precision;
mylat double precision;
lon1 float;
lon2 float;
lat1 float;
lat2 float;
BEGIN
raise notice 'help = %', $1; --just for testing
raise notice 'myvar = %', myvar; --again for testing
select cr.last_known_longitude, cr.last_known_latitude into mylon, mylat from current_reg as cr where userid = myvar;
lon1 = mylon - 1 / abs(cos(radians(mylat))*69);
lon2 = mylon + 1/abs(cos(radians(mylat))*69);
lat1 = mylat - (1/69);
lat2 = mylat + (1/69);
create view myview as
select destination.userid, 3956*2*asin(sqrt(power(sin((origin.last_known_latitude-destination.last_known_latitude)*pi()/180 / 2), 2) + cos(origin.last_known_latitude* pi()/180) * cos(destination.last_known_latitude* pi()/180) * power(sin((origin.last_known_longitude-destination.last_known_longitude) * pi()/180 /2), 2))) as distance
from current_reg destination, current_reg origin
where origin.userid = myvar
and destination.last_known_longitude between lon1 and lon2
and destination.last_known_latitude between lat1 and lat2
having distance < 1 order by distance limit 5;
return 0;
END;
$$ LANGUAGE 'plpgsql';
Select testfunction(7);
其中current_reg是一个以userid,last_known_latitude,last_known_longitude为列的表。作为整数传递给函数的参数是用户的用户ID,我希望在其位置(纬度和经度)中找到一英里范围内的用户。
我收到以下错误:
NOTICE: help = 7
NOTICE: myvar = 7
ERROR: column "myvar" does not exist
LINE 1: ...ination, current_reg origin where origin.userid = myvar and de...
^
QUERY: create view myview as select destination.userid, 3956*2*asin(sqrt(power(sin((origin.last_known_latitude-destination.last_known_latitude)*pi()/180 / 2), 2) + cos(origin.last_known_latitude* pi()/180) * cos(destination.last_known_latitude* pi()/180) * power(sin((origin.last_known_longitude-destination.last_known_longitude) * pi()/180 /2), 2))) as distance from current_reg destination, current_reg origin where origin.userid = myvar and destination.last_known_longitude between lon1 and lon2 and destination.last_known_latitude between lat1 and lat2 having distance < 1 order by distance limit 5
CONTEXT: PL/pgSQL function testfunction(integer) line 19 at SQL statement
********** Error **********
ERROR: column "myvar" does not exist
SQL state: 42703
Context: PL/pgSQL function testfunction(integer) line 19 at SQL statement
自&#34; myvar&#34;是一个变量,它为什么期望它成为一个列?
在@a_horse_with_no_name的帮助下,这是修改后的代码
CREATE OR REPLACE FUNCTION testfunction (p_userid integer)
RETURNS table (userid integer, distance float)
AS
$$
DECLARE
mylon double precision;
mylat double precision;
lon1 float;
lon2 float;
lat1 float;
lat2 float;
BEGIN
raise notice 'help = %', p_userid; --just for testing
select cr.last_known_longitude, cr.last_known_latitude
into mylon, mylat
from current_reg as cr
where cr.userid = p_userid;
lon1 = mylon - 1 / abs(cos(radians(mylat))*69);
lon2 = mylon + 1/abs(cos(radians(mylat))*69);
lat1 = mylat - (1/69);
lat2 = mylat + (1/69);
return query
select destination.userid,
3956*2*asin(sqrt(power(sin((origin.last_known_latitude-destination.last_known_latitude)*pi()/180 / 2), 2) + cos(origin.last_known_latitude* pi()/180) * cos(destination.last_known_latitude* pi()/180) * power(sin((origin.last_known_longitude-destination.last_known_longitude) * pi()/180 /2), 2)))
as distance
from current_reg as destination JOIN current_reg as origin
where origin.userid = p_userid
and destination.last_known_longitude between lon1 and lon2
and destination.last_known_latitude between lat1 and lat2
having distance < 1
order by distance limit 5;
END;
$$ LANGUAGE plpgsql;
Select userid from testfunction(4);
我现在收到以下错误: 错误:语法错误在或附近&#34;其中&#34; 第29行:其中origin.userid = p_userid ^
答案 0 :(得分:3)
如果要返回查询结果,则需要在PL / pgSQL中使用return query
。您不能在这样的DDL语句中使用变量。为每次调用函数创建一个视图是一个非常糟糕的主意。而且还有更多:第二次调用时,你的函数会失败,因为视图已经存在。
根据你所写的内容,我认为你想要这样的东西:
CREATE OR REPLACE FUNCTION testfunction (p_userid integer)
RETURNS table (userid integer, distance float)
AS
$$
DECLARE
mylon double precision;
mylat double precision;
lon1 float;
lon2 float;
lat1 float;
lat2 float;
BEGIN
raise notice 'help = %', p_userid; --just for testing
select cr.last_known_longitude, cr.last_known_latitude
into mylon, mylat
from current_reg as cr
where userid = p_userid;
lon1 = mylon - 1 / abs(cos(radians(mylat))*69);
lon2 = mylon + 1/abs(cos(radians(mylat))*69);
lat1 = mylat - (1/69);
lat2 = mylat + (1/69);
return query
select destination.userid,
3956*2*asin(sqrt(power(sin((origin.last_known_latitude-destination.last_known_latitude)*pi()/180 / 2), 2) + cos(origin.last_known_latitude* pi()/180) * cos(destination.last_known_latitude* pi()/180) * power(sin((origin.last_known_longitude-destination.last_known_longitude) * pi()/180 /2), 2))) as distance
from current_reg destination,
current_reg origin
where origin.userid = p_userid
and destination.last_known_longitude between lon1 and lon2
and destination.last_known_latitude between lat1 and lat2
having distance < 1
order by distance limit 5;
END;
$$ LANGUAGE plpgsql;
returns table
。 return query
。要获得该函数的结果,请使用:
select *
from testfunction(1);
请注意,该函数位于FROM
子句中,而不在SELECT
列表中!
仍然看起来可疑的事情:
from current_reg destination, current_reg origin
在两者之间创建交叉加入。使用显式JOIN
的另一个好例子比where子句having
但未使用任何聚合