问题
Is there a way to accomplish this?
INSERT IGNORE INTO some_table (one,two,three) VALUES(1,2,3)
ON DUPLICATE KEY (INSERT INTO audit_table VALUES(NOW(),'Duplicate key ignored')
I really don't want to use PHP for this :(
Thanks!
回答1:
If you want to consider using a stored procedure, you can use a DECLARE CONTINUE HANDLER. Here's an example:
CREATE TABLE users (
username VARCHAR(30),
first_name VARCHAR(30),
last_name VARCHAR(30),
PRIMARY KEY (username)
);
CREATE TABLE audit_table (timestamp datetime, description varchar(255));
DELIMITER $$
CREATE PROCEDURE add_user
(in_username VARCHAR(30),
in_first_name VARCHAR(30),
in_last_name VARCHAR(30))
MODIFIES SQL DATA
BEGIN
DECLARE duplicate_key INT DEFAULT 0;
BEGIN
DECLARE EXIT HANDLER FOR 1062 SET duplicate_key = 1;
INSERT INTO users (username, first_name, last_name)
VALUES (in_username, in_first_name, in_last_name);
END;
IF duplicate_key = 1 THEN
INSERT INTO audit_table VALUES(NOW(), 'Duplicate key ignored');
END IF;
END$$
DELIMITER ;
Let's add some data, trying to insert a duplicate key:
CALL add_user('userA', 'Bob', 'Smith');
CALL add_user('userB', 'Paul', 'Green');
CALL add_user('userA', 'Jack', 'Brown');
Result:
SELECT * FROM users;
+----------+------------+-----------+
| username | first_name | last_name |
+----------+------------+-----------+
| userA | Bob | Smith |
| userB | Paul | Green |
+----------+------------+-----------+
2 rows in set (0.00 sec)
SELECT * FROM audit_table;
+---------------------+-----------------------+
| timestamp | description |
+---------------------+-----------------------+
| 2010-10-07 20:17:35 | Duplicate key ignored |
+---------------------+-----------------------+
1 row in set (0.00 sec)
If auditing is important on a database level, you may want to grant EXECUTE
permissions only so that your database users can only call stored procedures.
回答2:
ON DUPLICATE KEY is used to update the row, not to insert in another table, you have to use two queries according the first's result.
EDIT : you can use a trigger too
回答3:
Not sure if this would work but look into IF statement for MySQL
Pseudo-code
IF(SELECT id_key_index FROM tbl WHERE id_key_index = $index) THEN (UPDATE SECOND TBL);
ELSE
INSERT ...
END IF;
来源:https://stackoverflow.com/questions/3884344/mysql-on-duplicate-key-insert-into-an-audit-or-log-table