Postgresql:错误:列";“梅瓦尔”;不存在,而myvar是一个变量
我试图在plsql中创建一个函数,以查找从作为参数传递到函数的用户1英里范围内的5个用户 这是我的代码:Postgresql:错误:列";“梅瓦尔”;不存在,而myvar是一个变量,postgresql,postgresql-9.4,Postgresql,Postgresql 9.4,我试图在plsql中创建一个函数,以查找从作为参数传递到函数的用户1英里范围内的5个用户 这是我的代码: CREATE OR REPLACE FUNCTION testfunction (integer) RETURNS integer as $$ DECLARE myvar integer := $1; mylon double precision; mylat double precision; lon1 float; lon2
CREATE OR REPLACE FUNCTION testfunction (integer) RETURNS integer as $$
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;
$$ LANGUAGE 'plpgsql';
Select testfunction(7);
创建或替换函数testfunction(integer)返回整数作为$$
声明
myvar整数:=$1;
迈伦双精度;
迈拉双精度;
lon1浮球;
lon2浮球;
lat1浮子;
lat2浮子;
开始
提出通知“帮助=%”,1美元--只是为了测试
提出通知“myvar=%”,myvar--再次进行测试
选择cr.last_known_经度,cr.last_known_纬度到mylon,mylat从当前_注册为cr,其中userid=myvar;
lon1=mylon-1/abs(cos(弧度(mylat))*69);
lon2=mylon+1/abs(cos(弧度(mylat))*69);
lat1=mylat-(1/69);
lat2=迈拉特+(1/69);
创建视图myview作为
选择destination.userid,3956*2*asin(sqrt(功率(sin((原点、最后一个已知纬度)destination.最后一个已知纬度)*pi()/180/2),2)+cos(原点、最后一个已知纬度*pi()/180)*cos(destination.最后一个已知纬度*pi()/180)*功率(sin sin(原点、最后一个已知经度、最后一个已知纬度)*pi()/180/2)作为距离
从当前\u reg目的地、当前\u reg来源
其中origin.userid=myvar
和destination.last_known_lon1和lon2之间的经度
和目的地。lat1和lat2之间的最后已知纬度
距离小于1个数量级,距离限制为5;
返回0;
结束;
$$语言“plpgsql”;
选择testfunction(7);
其中,current_reg是一个以userid、last_known_纬度、last_known_经度为列的表。作为整数传递给函数的参数是我希望从其位置(纬度和经度)查找英里范围内用户的用户ID
我得到以下错误:
NOTICE: help = 7
NOTICE: myvar = 7
ERROR: column "myvar" does not exist
LINE 1: ...ination, current_reg origin where origin.userid = myvar and de...
^
QUERY: create view myview as select destination.userid, 3956*2*asin(sqrt(power(sin((origin.last_known_latitude-destination.last_known_latitude)*pi()/180 / 2), 2) + cos(origin.last_known_latitude* pi()/180) * cos(destination.last_known_latitude* pi()/180) * power(sin((origin.last_known_longitude-destination.last_known_longitude) * pi()/180 /2), 2))) as distance from current_reg destination, current_reg origin where origin.userid = myvar and destination.last_known_longitude between lon1 and lon2 and destination.last_known_latitude between lat1 and lat2 having distance < 1 order by distance limit 5
CONTEXT: PL/pgSQL function testfunction(integer) line 19 at SQL statement
********** Error **********
ERROR: column "myvar" does not exist
SQL state: 42703
Context: PL/pgSQL function testfunction(integer) line 19 at SQL statement
注意:help=7
注意:myvar=7
错误:“myvar”列不存在
第1行:…初始化,当前_reg origin,其中origin.userid=myvar和de。。。
^
查询:创建视图myview作为select destination.userid,3956*2*asin(sqrt(功率(sin((原点.上次已知纬度-destination.last已知纬度)*pi()/180/2),2)+cos(原点.上次已知纬度*pi()/180)*cos(destination.last已知纬度*pi()/180)*功率(sin sin(原点.上次已知经度-destination.last已知经度)*pi()/180/2))作为与当前_reg目的地的距离,当前_reg原点,其中origin.userid=myvar和destination.last_known_lon1和lon2之间的经度和destination.last_known_lation lat1和lat2之间的纬度,距离小于1,距离限制为5
上下文:PL/pgSQL函数testfunction(整数)SQL语句第19行
**********错误**********
错误:“myvar”列不存在
SQL状态:42703
上下文:PL/pgSQL函数testfunction(整数)SQL语句第19行
既然“myvar”是一个变量,为什么它希望它是一个列呢
在@a_horse_(没有名字)的帮助下,这是修改过的代码
CREATE OR REPLACE FUNCTION testfunction (p_userid integer)
RETURNS table (userid integer, distance float)
AS
$$
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;
$$ LANGUAGE plpgsql;
Select userid from testfunction(4);
创建或替换函数testfunction(p_userid integer)
返回表(用户标识整数、距离浮点)
作为
$$
声明
迈伦双精度;
迈拉双精度;
lon1浮球;
lon2浮球;
lat1浮子;
lat2浮子;
开始
发出通知“help=%”,p_userid--只是为了测试
选择cr.last\u known\u经度、cr.last\u known\u纬度
迈伦,迈拉
从当前注册为cr
其中cr.userid=p_userid;
lon1=mylon-1/abs(cos(弧度(mylat))*69);
lon2=mylon+1/abs(cos(弧度(mylat))*69);
lat1=mylat-(1/69);
lat2=迈拉特+(1/69);
返回查询
选择destination.userid,
3956*2*asin(sqrt(功率(sin((原点、上次已知纬度-目的地、上次已知纬度)*pi()/180/2),2)+cos(原点、上次已知纬度*pi()/180)*cos(目的地、上次已知纬度*pi()/180)*功率(sin((原点、上次已知经度-目的地、上次已知经度)*pi()/180/2)))
作为距离
从当前\u reg作为目标加入当前\u reg作为源
其中origin.userid=p_userid
和destination.last_known_lon1和lon2之间的经度
和目的地。lat1和lat2之间的最后已知纬度
距离小于1的
按距离限制订购5件;
结束;
$$语言plpgsql;
从testfunction(4)中选择userid;
我现在得到以下错误:
错误:在“where”处或附近出现语法错误
第29行:其中origin.userid=p_userid
^如果要返回查询结果,需要在PL/pgSQL中使用
返回查询。不能在DDL语句中使用这样的变量。为函数的每次调用创建一个视图是一个非常糟糕的主意。更重要的是:第二次调用时,函数将失败,因为视图已经存在
根据你所写的,我认为你想要这样的东西:
CREATE OR REPLACE FUNCTION testfunction (p_userid integer)
RETURNS table (userid integer, distance float)
AS
$$
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;
注意该函数位于FROM
子句中,而不是SELECT
列表中
看起来仍然可疑的事情:
您没有从当前的\u reg目的地连接这两个表,当前的\u reg origin
会在这两个表之间创建一个交叉连接。另一个很好的例子是,为什么在where子句中使用显式的连接
优于旧的隐式连接
具有
,但未使用任何聚合如果要返回查询结果,需要在PL/pgSQL中使用
returnquery
。不能在DDL语句中使用这样的变量。为函数的每次调用创建一个视图是一个非常糟糕的主意。更重要的是:第二次调用时,函数将失败,因为视图已经存在
select *
from testfunction(1);