I've searched around but didn't find if it's possible.
I've this MySQL query:
INSERT INTO table (id,a,b,c,d,e,f,g) VALUES (1,2,3,4,5,6,7,8)
Field id has a "unique index", so there can't be two of them. Now if the same id is already present in the database, I'd like to update it. But do I really have to specify all these field again, like:
INSERT INTO table (id,a,b,c,d,e,f,g) VALUES (1,2,3,4,5,6,7,8)
ON DUPLICATE KEY UPDATE a=2,b=3,c=4,d=5,e=6,f=7,g=8
Or:
INSERT INTO table (id,a,b,c,d,e,f,g) VALUES (1,2,3,4,5,6,7,8)
ON DUPLICATE KEY UPDATE a=VALUES(a),b=VALUES(b),c=VALUES(c),d=VALUES(d),e=VALUES(e),f=VALUES(f),g=VALUES(g)
I've specified everything already in the insert...
A extra note, I'd like to use the work around to get the ID to!
id=LAST_INSERT_ID(id)
I hope somebody can tell me what the most efficient way is.
The UPDATE statement is given so that older fields can be updated to new value. If your older values are the same as your new ones, why would you need to update it in any case?
For eg. if your columns a to g are already set as 2 to 8; there would be no need to re-update it.
Alternatively, you can use:
INSERT INTO table (id,a,b,c,d,e,f,g)
VALUES (1,2,3,4,5,6,7,8)
ON DUPLICATE KEY
UPDATE a=a, b=b, c=c, d=d, e=e, f=f, g=g;
To get the id from LAST_INSERT_ID; you need to specify the backend app you're using for the same.
For LuaSQL, a conn:getlastautoid() fetches the value.
There is a MySQL specific extension to SQL that may be what you want - REPLACE INTO
However it does not work quite the same as 'ON DUPLICATE UPDATE'
It deletes the old row that clashes with the new row and then inserts the new row. So long as you don't have a primary key on the table that would be fine, but if you do, then if any other table references that primary key
You can't reference the values in the old rows so you can't do an equivalent of
INSERT INTO mytable (id, a, b, c) values ( 1, 2, 3, 4)
ON DUPLICATE KEY UPDATE
id=1, a=2, b=3, c=c + 1;
I'd like to use the work around to get the ID to!
That should work — last_insert_id() should have the correct value so long as your primary key is auto-incrementing.
However as I said, if you actually use that primary key in other tables, REPLACE INTO probably won't be acceptable to you, as it deletes the old row that clashed via the unique key.
Someone else suggested before you can reduce some typing by doing:
INSERT INTO `tableName` (`a`,`b`,`c`) VALUES (1,2,3)
ON DUPLICATE KEY UPDATE `a`=VALUES(`a`), `b`=VALUES(`b`), `c`=VALUES(`c`);
There is no other way, I have to specify everything twice. First for the insert, second in the update case.
Here is a solution to your problem:
I've tried to solve problem like yours & I want to suggest to test from simple aspect.
Follow these steps: Learn from simple solution.
Step 1: Create a table schema using this SQL Query:
CREATE TABLE IF NOT EXISTS `user` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(30) NOT NULL,
`password` varchar(32) NOT NULL,
`status` tinyint(1) DEFAULT '0',
PRIMARY KEY (`id`),
UNIQUE KEY `no_duplicate` (`username`,`password`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1;
Step 2: Create an index of two columns to prevent duplicate data using following SQL Query:
ALTER TABLE `user` ADD INDEX no_duplicate (`username`, `password`);
or, Create an index of two column from GUI as follows:
Step 3: Update if exist, insert if not using following queries:
INSERT INTO `user`(`username`, `password`) VALUES ('ersks','Nepal') ON DUPLICATE KEY UPDATE `username`='master',`password`='Nepal';
INSERT INTO `user`(`username`, `password`) VALUES ('master','Nepal') ON DUPLICATE KEY UPDATE `username`='ersks',`password`='Nepal';
Just in case you are able to utilize a scripting language to prepare your SQL queries, you could reuse field=value pairs by using SET instead of (a,b,c) VALUES(a,b,c).
An example with PHP:
$pairs = "a=$a,b=$b,c=$c";
$query = "INSERT INTO $table SET $pairs ON DUPLICATE KEY UPDATE $pairs";
Example table:
CREATE TABLE IF NOT EXISTS `tester` (
`a` int(11) NOT NULL,
`b` varchar(50) NOT NULL,
`c` text NOT NULL,
UNIQUE KEY `a` (`a`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
I know it's late, but i hope someone will be helped of this answer
INSERT INTO t1 (a,b,c) VALUES (1,2,3),(4,5,6)
ON DUPLICATE KEY UPDATE c=VALUES(a)+VALUES(b);
You can read the tutorial below here :
https://mariadb.com/kb/en/library/insert-on-duplicate-key-update/
http://www.mysqltutorial.org/mysql-insert-or-update-on-duplicate-key-update/
You may want to consider using REPLACE INTO syntax, but be warned, upon duplicate PRIMARY / UNIQUE key, it DELETES the row and INSERTS a new one.
You won't need to re-specify all the fields. However, you should consider the possible performance reduction (depends on your table design).
Caveats:
If you have AUTO_INCREMENT primary key, it will be given a new one
Indexes will probably need to be updated
With MySQL v8.0.19 and above you can do this:
mysql doc
INSERT INTO mytable(fielda, fieldb, fieldc)
VALUES("2022-01-01", 97, "hello")
AS NEW(newfielda, newfieldb, newfieldc)
ON DUPLICATE KEY UPDATE
fielda=newfielda,
fieldb=newfieldb,
fieldc=newfieldc;
SIDENOTE: Also if you want a conditional in the on duplicate key update part there is a twist in MySQL. If you update fielda as the first argument and include it inside the IF clause for fieldb it will already be updated to the new value! Move it to the end or alike. Let's say fielda is a date like in the example and you want to update only if the date is newer than the previous:
INSERT INTO mytable(fielda, fieldb)
VALUES("2022-01-01", 97)
AS NEW(newfielda, newfieldb, newfieldc)
ON DUPLICATE KEY UPDATE
fielda=IF(fielda<STR_TO_DATE(newfielda,'%Y-%m-%d %H:%i:%s'),newfielda,fielda),
fieldb=IF(fielda<STR_TO_DATE(newfielda,'%Y-%m-%d %H:%i:%s'),newfieldb,fieldb);
in this case fieldb would never be updated because of the <! you need to move the update of fielda below it or check with <= or =...!
INSERT INTO mytable(fielda, fieldb)
VALUES("2022-01-01", 97)
AS NEW(newfielda, newfieldb, newfieldc)
ON DUPLICATE KEY UPDATE
fielda=IF(fielda<STR_TO_DATE(newfielda,'%Y-%m-%d %H:%i:%s'),newfielda,fielda),
fieldb=IF(fielda=STR_TO_DATE(newfielda,'%Y-%m-%d %H:%i:%s'),newfieldb,fieldb);
This works as expected with using = since fielda is already updated to its new value before reaching the if clause of fieldb... Personally i like <= the most in such a case if you ever rearrange the statement...
you can use insert ignore for such case, it will ignore if it gets duplicate records
INSERT IGNORE
... ; -- without ON DUPLICATE KEY
I'm having a bit of a strange problem. I'm trying to add a foreign key to one table that references another, but it is failing for some reason. With my limited knowledge of MySQL, the only thing that could possibly be suspect is that there is a foreign key on a different table referencing the one I am trying to reference.
I've done a SHOW CREATE TABLE query on both tables, sourcecodes_tags is the table with the foreign key, sourcecodes is the referenced table.
CREATE TABLE `sourcecodes` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) unsigned NOT NULL,
`language_id` int(11) unsigned NOT NULL,
`category_id` int(11) unsigned NOT NULL,
`title` varchar(40) CHARACTER SET utf8 NOT NULL,
`description` text CHARACTER SET utf8 NOT NULL,
`views` int(11) unsigned NOT NULL,
`downloads` int(11) unsigned NOT NULL,
`time_posted` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`),
KEY `language_id` (`language_id`),
KEY `category_id` (`category_id`),
CONSTRAINT `sourcecodes_ibfk_3` FOREIGN KEY (`language_id`) REFERENCES `languages` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `sourcecodes_ibfk_2` FOREIGN KEY (`category_id`) REFERENCES `categories` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1
CREATE TABLE `sourcecodes_tags` (
`sourcecode_id` int(11) unsigned NOT NULL,
`tag_id` int(11) unsigned NOT NULL,
KEY `sourcecode_id` (`sourcecode_id`),
KEY `tag_id` (`tag_id`),
CONSTRAINT `sourcecodes_tags_ibfk_1` FOREIGN KEY (`tag_id`) REFERENCES `tags` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1
This is the code that generates the error:
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
Quite likely your sourcecodes_tags table contains sourcecode_id values that no longer exists in your sourcecodes table. You have to get rid of those first.
Here's a query that can find those IDs:
SELECT DISTINCT sourcecode_id FROM
sourcecodes_tags tags LEFT JOIN sourcecodes sc ON tags.sourcecode_id=sc.id
WHERE sc.id IS NULL;
I had the same issue with my MySQL database but finally, I got a solution which worked for me.
Since in my table everything was fine from the mysql point of view(both tables should use InnoDB engine and the datatype of each column should be of the same type which takes part in foreign key constraint).
The only thing that I did was to disable the foreign key check and later on enabled it after performing the foreign key operation.
Steps that I took:
SET foreign_key_checks = 0;
alter table tblUsedDestination add constraint f_operatorId foreign key(iOperatorId) references tblOperators (iOperatorId); Query
OK, 8 rows affected (0.23 sec) Records: 8 Duplicates: 0 Warnings: 0
SET foreign_key_checks = 1;
Use NOT IN to find where constraints are constraining:
SELECT column FROM table WHERE column NOT IN
(SELECT intended_foreign_key FROM another_table)
so, more specifically:
SELECT sourcecode_id FROM sourcecodes_tags WHERE sourcecode_id NOT IN
(SELECT id FROM sourcecodes)
EDIT: IN and NOT IN operators are known to be much faster than the JOIN operators, as well as much easier to construct, and repeat.
Truncate the tables and then try adding the FK Constraint.
I know this solution is a bit awkward but it does work 100%. But I agree that this is not an ideal solution to deal with problem, but I hope it helps.
For me, this problem was a little different and super easy to check and solve.
You must ensure BOTH of your tables are InnoDB. If one of the tables, namely the reference table is a MyISAM, the constraint will fail.
SHOW TABLE STATUS WHERE Name = 't1';
ALTER TABLE t1 ENGINE=InnoDB;
This also happens when setting a foreign key to parent.id to child.column if the child.column has a value of 0 already and no parent.id value is 0
You would need to ensure that each child.column is NULL or has value that exists in parent.id
And now that I read the statement nos wrote, that's what he is validating.
I had the same problem today. I tested for four things, some of them already mentioned here:
Are there any values in your child column that don't exist in the parent column (besides NULL, if the child column is nullable)
Do child and parent columns have the same datatype?
Is there an index on the parent column you are referencing? MySQL seems to require this for performance reasons (http://dev.mysql.com/doc/refman/5.5/en/create-table-foreign-keys.html)
And this one solved it for me: Do both tables have identical collation?
I had one table in UTF-8 and the other in iso-something. That didn't work. After changing the iso-table to UTF-8 collation the constraints could be added without problems. In my case, phpMyAdmin didn't even show the child table in iso-encoding in the dropdown for creating the foreign key constraint.
It seems there is some invalid value for the column line 0 that is not a valid foreign key so MySQL cannot set a foreign key constraint for it.
You can follow these steps:
Drop the column which you have tried to set FK constraint for.
Add it again and set its default value as NULL.
Try to set a foreign key constraint for it again.
I'd the same problem, I checked rows of my tables and found there was some incompatibility with the value of fields that I wanted to define a foreign key. I corrected those value, tried again and the problem was solved.
I end up delete all the data in my table, and run alter again. It works. Not the brilliant one, but it save a lot time, especially your application is still in development stage without any customer data.
try this
SET foreign_key_checks = 0;
ALTER TABLE sourcecodes_tags ADD FOREIGN KEY (sourcecode_id) REFERENCES sourcecodes (id) ON DELETE CASCADE ON UPDATE CASCADE
SET foreign_key_checks = 1;
I had this exact same problem about three different times. In each instance it was because one (or more) of my records did not conform to the new foreign key. You may want to update your existing records to follow the syntax constraints of the foreign key before trying to add the key itself. The following example should generally isolate the problem records:
SELECT * FROM (tablename)
WHERE (candidate key) <> (proposed foreign key value)
AND (candidate key) <> (next proposed foreign key value)
repeat AND (candidate key) <> (next proposed foreign key value) within your query for each value in the foreign key.
If you have a ton of records this can be difficult, but if your table is reasonably small it shouldn't take too long. I'm not super amazing in SQL syntax, but this has always isolated the issue for me.
Empty both your tables' data and run the command. It will work.
I was getting this error when using Laravel and eloquent, trying to make a foreign key link would cause a 1452. The problem was lack of data in the linked table.
Please see here for an example: http://mstd.eu/index.php/2016/12/02/laravel-eloquent-integrity-constraint-violation-1452-foreign-key-constraint/
You just need to answer one question:
Is your table already storing data? (Especially the table included foreign key.)
If the answer is yes, then the only thing you need to do is to delete all the records, then you are free to add any foreign key to your table.
Delete instruction: From child(which include foreign key table) to parent table.
The reason you cannot add in foreign key after data entries is due to the table inconsistency, how are you going to deal with a new foreign key on the former data-filled the table?
If the answer is no, then follow other instructions.
I was readying this solutions and this example may help.
My database have two tables (email and credit_card) with primary keys for their IDs. Another table (client) refers to this tables IDs as foreign keys. I have a reason to have the email apart from the client data.
First I insert the row data for the referenced tables (email, credit_card) then you get the ID for each, those IDs are needed in the third table (client).
If you don't insert first the rows in the referenced tables, MySQL wont be able to make the correspondences when you insert a new row in the third table that reference the foreign keys.
If you first insert the referenced rows for the referenced tables, then the row that refers to foreign keys, no error occurs.
Hope this helps.
Make sure the value is in the other table otherwise you will get this error, in the assigned corresponding column.
So if it is assigned column is assigned to a row id of another table , make sure there is a row that is in the table otherwise this error will appear.
you can try this exapmple
START TRANSACTION;
SET foreign_key_checks = 0;
ALTER TABLE `job_definers` ADD CONSTRAINT `job_cities_foreign` FOREIGN KEY
(`job_cities`) REFERENCES `drop_down_lists`(`id`) ON DELETE CASCADE ON UPDATE CASCADE;
SET foreign_key_checks = 1;
COMMIT;
Note : if you are using phpmyadmin just uncheck Enable foreign key checks
as example
hope this soloution fix your problem :)
UPDATE sourcecodes_tags
SET sourcecode_id = NULL
WHERE sourcecode_id NOT IN (
SELECT id FROM sourcecodes);
should help to get rid of those IDs. Or if null is not allowed in sourcecode_id, then remove those rows or add those missing values to the sourcecodes table.
I had the same problem and found solution, placing NULL instead of NOT NULL on foreign key column. Here is a query:
ALTER TABLE `db`.`table1`
ADD COLUMN `col_table2_fk` INT UNSIGNED NULL,
ADD INDEX `col_table2_fk_idx` (`col_table2_fk` ASC),
ADD CONSTRAINT `col_table2_fk1`
FOREIGN KEY (`col_table2_fk`)
REFERENCES `db`.`table2` (`table2_id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
MySQL has executed this query!
In my case, I created a new table with the same structure, created the relationships with the other tables, then extracted the data in CSV from the old table that has the problem, then imported the CSV to the new table and disabled foreign key checking and disabled import interruption, all my data are inserted to the new table that has no problem successfully, then deleted the old table.
It worked for me.