我有以下代码可以使用:
UPDATE backup_factura
SET tipo = CASE
WHEN total_fact <=100 THEN 'X'
WHEN total_fact <=200 THEN 'Y'
ELSE 'Z'
END
RETURNING *;
我的老师让我把它转换成一个功能。 我尝试过以下方法:
CREATE Function sp_test_case () returns void as $$
BEGIN
UPDATE backup_factura
SET tipo = CASE
WHEN total_fact <=100 THEN 'X'
WHEN total_fact <=200 THEN 'Y'
ELSE 'Z'
END
RETURNING *;
RETURN;
END;
$$ LANGUAGE plpgsql;
但是当我执行该功能时,我得到:
********** Error **********
ERROR: query has no destination for result data
SQL state: 42601
Context: PL / pgSQL sp_test_case () function in line 4 SQL statement
我也试过更复杂的方法:
CREATE Function sp_test_case () returns void as $$
DECLARE
cont int=(Select MAX(id_fact)from backup_factura);
BEGIN
while cont>0
LOOP
UPDATE backup_factura
SET tipo= CASE
WHEN ((total_fact) <=100) THEN 'X'
WHEN ((total_fact) <=200) THEN 'Y'
ELSE 'Z'
END;
WHERE id_fact=cont;
cont:=cont-1;
END LOOP;
RETURN;
END;
$$ LANGUAGE plpgsql;
但我明白了:
********** Error **********
ERROR: syntax error at or near "WHERE"
SQL state: 42601
Character: 283
重点是显示如下:
答案 0 :(得分:1)
您可以使用return query
语法并返回setof
您的表格:
CREATE Function sp_test_case () RETURNS SETOF backup_factura AS $$
BEGIN
RETURN QUERY
UPDATE backup_factura
SET tipo = CASE
WHEN total_fact <=100 THEN 'X'
WHEN total_fact <=200 THEN 'Y'
ELSE 'Z'
END
RETURNING *;
END;
$$ LANGUAGE plpgsql;
答案 1 :(得分:1)
你可以像@Mureinik发布的那样修复你的plpgsql函数。或者,更好的是,使用更简单的SQL function:
CREATE FUNCTION sp_test_case ()
RETURNS SETOF backup_factura AS
$func$
UPDATE backup_factura
SET tipo = CASE WHEN total_fact <= 100 THEN 'X'
WHEN total_fact <= 200 THEN 'Y'
ELSE 'Z' END
RETURNING *;
$func$ LANGUAGE sql;
呼叫:
SELECT * FROM sp_test_case();