sql - 错误 : return type mismatch in function declared to return

Question

基于这些表:

create table f1_driver(
 code varchar(5) not null primary key,
 name varchar(10),
 surname varchar(20),
 dateofbirth date,
 debut integer,
 countryoforigin varchar(20),
 points integer
);

create table f1_results (
 drivercode varchar(5) not null references f1_driver,
 circuitcode varchar(5) not null references f1_circuit,
 racedate date,
 raceposition integer,
 grid integer,
 primary key (drivercode, circuitcode,  racedate)
);

我想创建一个函数，用户将提供 circuitcode，该函数将返回驱动程序的 name 和 surname在这个特定的电路中，raceposition 比 grid 更好。

我这样写:

CREATE FUNCTION get(character) RETURNS SETOF f1_driver AS
$$
SELECT  D.name, D.surname
FROM f1_driver D,f1_results R
WHERE R.circuitcode = $1
AND D.code=R.drivercode
AND R.raceposition<grid ;
$$ LANGUAGE SQL;

我有这个错误:

ERROR:  return type mismatch in function declared to return f1_driver
DETAIL:  Final statement returns too few columns.
CONTEXT:  SQL function "get"

Answer 1

行类型 f1_driver 与您实际返回的不匹配。使用 RETURNS TABLE提供匹配声明:

CREATE FUNCTION f_get_drivers(varchar)
  RETURNS TABLE(name varchar, surname varchar) AS
$func$
SELECT D.name, D.surname
FROM   f1_driver D
JOIN   f1_results R ON R.drivercode = D.code
WHERE  R.circuitcode = $1
AND    R.raceposition < R.grid;
$func$ LANGUAGE sql;

要点

• RETURNS 中声明的返回类型必须与实际返回的类型相匹配。

• character 是您输入参数的错误数据类型。使用 text 或 varchar。详情:

  • Any downsides of using data type "text" for storing strings?

  • Compare varchar with char

• 使用显式 JOIN 做同样的事情。这样更易于阅读和维护。