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In many situations, you may want to ensure that a record exists in a table without a conflicting entry. Essentially, you want to find and modify the current record if it exists or add a new record with the values you want if one is not already present. This is typically referred to as an upsert operation (a combination of insert and update).MySQL allows you to perform this action using the ON DUPLICATE KEY UPDATE clause to modify the INSERT command. In this guide, well cover how to use this construct to update the values of an entry if it exists or else add it as a new row in the table.The basic syntax for the insert or update operation looks like this:INSERT INTO my_table (column1, column2)Multiple columns can be provided after the ON DUPLICATE KEY UPDATE clause, each defining what the new value should be if theres a conflict with an existing record.To demonstrate this feature, well imagine a table called director with the following columns and populated data: name VARCHAR(200) NOT NULL,INSERT INTO director (name)The data within the table looks like this:+----+-------+-------------+ id | name | latest_film |+----+-------+-------------++----+-------+-------------+If you attempt to insert another row with an id column of 3, MySQL will notify you of a conflict with the existing row:INSERT INTO director (id, name) VALUES (3, susan);ERROR 1062 (23000): Duplicate entry 3 for key director.PRIMARYIf we anticipate this possibility and want to update the existing row with the new information, we can avoid this error. The ON DUPLICATE KEY UPDATE clause allows us to do this:INSERT INTO director (id, name) VALUES (3, susan)ON DUPLICATE KEY UPDATE name = susan;Query OK, 2 rows affected (0.00 sec)MySQL considers an ON DUPLICATE KEY UPDATE where an update occurs to the existing row as two rows affected. If no conflict had occurred and the new record had been added, it would instead show one row affected. If an existing record was found but the columns already had the correct value, no rows would be reported as affected.You can confirm the row has been updated with the new information by typing:+----+-------+-------------+ id | name | latest_film |+----+-------+-------------++----+-------+-------------+If you are attempting to insert or update multiple records at the same time, the value to set each column likely depends on which record or records conflicted. For instance if you are trying to insert four new rows, but the third row has an id column that conflicts with an existing record, you most likely want to update the existing row based on the data you had in mind for the third row.MySQL allows you to reference that proposed data using the VALUES() function. The function takes a column name as an argument and provides the value given in the INSERT portion of the statement.The basic syntax looks like this:INSERT INTO my_table (column1, column2) <column1> = VALUES(<column1>), <column2> = VALUES(<column2>);This tells MySQL to use the data that is associated with the conflicting row to update the values.To see how this works, lets go back to the director table from before:+----+-------+-------------+ id | name | latest_film |+----+-------+-------------++----+-------+-------------+Suppose we wanted to ensure that the following records exist in the table:+----+--------+-------------+ id | name | latest_film |+----+--------+-------------++----+--------+-------------+We could create an INSERT...ON DUPLICATE KEY UPDATE statement that looks like this:INSERT INTO director (id, name)MySQL would accept the statement, inserting two new rows and updating one where it conflicted with an existing record (there is already a record with an id of 2):Query OK, 4 rows affected, 1 warning (0.01 sec)Records: 3 Duplicates: 1 Warnings: 1If you view the tables data, you can see that the two new rows appear as expected and the values of the conflicting row have been updated with the appropriate new information:+----+--------+-------------+ id | name | latest_film |+----+--------+-------------++----+--------+-------------+MySQLs INSERT...ON DUPLICATE KEY UPDATE construct allows you to insert data while avoiding conflicts with existing records. Combined with the VALUES() function, you can use it to make contextual updates to records that already exist without issuing multiple statements. This powerful feature can help you minimize the amount of checks and conditional logic you have to use outside of your SQL statements. Applies to ✅ Open Source Edition ✅ Express Edition ✅ Professional Edition ✅ Enterprise Edition The MySQL database supports a very convenient way to INSERT or UPDATE a record. This is a non-standard extension to the SQL syntax, which is supported by jOOQ and emulated in other RDBMS, where this is possible (e.g. if they support the SQL standard MERGE statement). Here is an example how to use the ON DUPLICATE KEY UPDATE clause: Add a new author called Koontz with ID 3. If that ID is already present, update the authors namecreate.insertInto(AUTHOR, AUTHOR.ID, AUTHOR.LAST_NAME) .values(3, Koontz) .onDuplicateKeyUpdate() .set(AUTHOR.LAST_NAME, Koontz) .execute(); The MySQL database also supports an INSERT IGNORE INTO clause. This is supported by jOOQ using the more convenient SQL syntax variant of ON DUPLICATE KEY IGNORE: Add a new author called Koontz with ID 3. If that ID is already present, ignore the INSERT statementcreate.insertInto(AUTHOR, AUTHOR.ID, AUTHOR.LAST_NAME) .values(3, Koontz) .onDuplicateKeyIgnore() .execute(); If the underlying database doesnt have any way to ignore failing INSERT statements, (e.g. MySQL via INSERT IGNORE), jOOQ can emulate the statement using a MERGE statement, or using INSERT .. SELECT WHERE NOT EXISTS: Emulating IGNORE with MERGE The above jOOQ statement can be emulated with the following, equivalent SQL statement: MERGE INTO AUTHORUSING (SELECT 1 FROM DUAL)ON (AUTHOR.ID = 3)WHEN NOT MATCHED THEN INSERT (ID, LAST_NAME) VALUES (3, Koontz) The above jOOQ statement can be emulated with the following, equivalent SQL statement: INSERT INTO AUTHOR (ID, LAST_NAME)SELECT 3, KoontzWHERE NOT EXISTS ( SELECT 1 FROM AUTHOR WHERE AUTHOR.ID = 3)
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