我想寫一個選擇內部MySQL的功能,但總是得到一個NULL返回MySQL的選擇功能
CREATE FUNCTION test (i CHAR)
RETURNS CHAR
NOT DETERMINISTIC
BEGIN
DECLARE select_var CHAR;
SET select_var = (SELECT name FROM table WHERE id = i);
RETURN select_var;
END$$
mysql> SELECT test('1')$$
+-----------------+
| test('1') |
+-----------------+
| NULL |
+-----------------+
1 row in set, 1 warning (0.00 sec)
mysql>
mysql>
mysql> SHOW WARNINGS
-> $$
+---------+------+----------------------------------------+
| Level | Code | Message |
+---------+------+----------------------------------------+
| Warning | 1265 | Data truncated for column 'i' at row 1 |
+---------+------+----------------------------------------+
1 row in set (0.00 sec)
這可能是顯而易見的,但是在id = 1的表中有一行嗎? – Zohaib
當我自己運行select時,它返回一個結果。 – jdborg
爲什麼我是char而不是INT? –