S. *_*. N 18 sql postgresql if-statement
我是postgres的新手,我正在完成我的任务.我不得不创建一个只有1列的表,然后我得到了这个声明在pgadmin III上运行:
BEGIN;
INSERT INTO mytable VALUES (1);
SAVEPOINT savepoint1;
INSERT INTO mytable VALUES (2);
ROLLBACK TO SAVEPOINT savepoint1;
INSERT INTO mytable VALUES (3);
SAVEPOINT savepoint2;
INSERT INTO mytable VALUES (4);
INSERT INTO mytable VALUES (5);
SAVEPOINT savepoint3;
SELECT * FROM mytable;
--NOTE: You need to run this IF statement as PGScript
--(button next to the normal run button)
IF (CAST ((SELECT MAX(id) FROM mytable) AS INTEGER) = 4)
BEGIN
RELEASE SAVEPOINT savepoint2;
END
ELSE
BEGIN
INSERT INTO mytable VALUES(6);
END
--Run the next steps normally
SAVEPOINT savepoint2;
INSERT INTO mytable VALUES (7);
RELEASE SAVEPOINT savepoint2;
INSERT INTO mytable VALUES (8);
ROLLBACK TO savepoint2;
COMMIT;
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当我运行这个时,我收到此错误:语法错误在"IF"或附近
我已经看过这个38.6.2.Conditionals38.6.2.条件,我不太了解这一点,我是否需要更改查询
IF (CAST ((SELECT MAX(id) FROM mytable) AS INTEGER) = 4) THEN
BEGiN
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然后当它结束时我应该结束它:
END IF
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毕竟为什么会出错?
mu *_*ort 46
IF和其他PL/pgSQL功能仅在PL/pgSQL函数中可用.如果要使用,则需要将代码包装在函数中IF.如果你使用9.0+,那么你可以DO用来编写内联函数:
do $$
begin
-- code goes here
end
$$
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如果您使用的是PostgreSQL的早期版本,则必须编写包含代码的命名函数,然后执行该函数.
不是OP的答案,但可能是一些最终到达这里的人的答案(比如我自己):如果您在 BEGIN-END 块中声明变量,您将得到相同的语法错误。
所以这是错误的:
DO $$
BEGIN
DECLARE my_var VARCHAR(50) := 'foo';
IF my_var IS NULL THEN
--some logic
END IF;
END;
$$
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这应该可以修复它:
DO $$
DECLARE my_var VARCHAR(50) := 'foo';
BEGIN
IF my_var IS NULL THEN
--some logic
END IF;
END;
$$
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