ERROR 1452: Cannot add or update a child row: a foreign key constraint fails

Question

I have created tables in MySQL Workbench as shown below :

ORDRE table:

CREATE TABLE Ordre (
  OrdreID   INT NOT NULL,
  OrdreDato DATE DEFAULT NULL,
  KundeID   INT  DEFAULT NULL,
  CONSTRAINT Ordre_pk PRIMARY KEY (OrdreID),
  CONSTRAINT Ordre_fk FOREIGN KEY (KundeID) REFERENCES Kunde (KundeID)
)
  ENGINE = InnoDB;

PRODUKT table:

CREATE TABLE Produkt (
  ProduktID          INT NOT NULL,
  ProduktBeskrivelse VARCHAR(100) DEFAULT NULL,
  ProduktFarge       VARCHAR(20)  DEFAULT NULL,
  Enhetpris          INT          DEFAULT NULL,
  CONSTRAINT Produkt_pk PRIMARY KEY (ProduktID)
)
  ENGINE = InnoDB;

and ORDRELINJE table:

CREATE TABLE Ordrelinje (
  Ordre         INT NOT NULL,
  Produkt       INT NOT NULL,
  AntallBestilt INT DEFAULT NULL,
  CONSTRAINT Ordrelinje_pk PRIMARY KEY (Ordre, Produkt),
  CONSTRAINT Ordrelinje_fk FOREIGN KEY (Ordre) REFERENCES Ordre (OrdreID),
  CONSTRAINT Ordrelinje_fk1 FOREIGN KEY (Produkt) REFERENCES Produkt (ProduktID)
)
  ENGINE = InnoDB;

so when I try to insert values into ORDRELINJE table i get:

Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails (srdjank.Ordrelinje, CONSTRAINT Ordrelinje_fk FOREIGN KEY (Ordre) REFERENCES Ordre (OrdreID))

I've seen the other posts on this topic, but no luck. Am I overseeing something or any idea what to do?

possible duplicate of [Mysql error 1452 - Cannot add or update a child row: a foreign key constraint fails](http://stackoverflow.com/questions/1253459/mysql-error-1452-cannot-add-or-update-a-child-row-a-foreign-key-constraint-fa) — Sagar Naliyapara, Jul 16 '15 at 05:07
You need to empty your table before you add this foreign key — Naqvi, Mar 13 '23 at 07:27

Rahul · Answer 1 · 2020-05-23T21:00:04.697

293

Taken from Using FOREIGN KEY Constraints

Foreign key relationships involve a parent table that holds the central data values, and a child table with identical values pointing back to its parent. The FOREIGN KEY clause is specified in the child table.

It will reject any INSERT or UPDATE operation that attempts to create a foreign key value in a child table if there is no a matching candidate key value in the parent table.

So your error Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails essentially means that, you are trying to add a row to your Ordrelinje table for which no matching row (OrderID) is present in Ordre table.

You must first insert the row to your Ordre table.

edited May 23 '20 at 21:00

answered Feb 09 '14 at 15:18

Rahul

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Or can we drop the foreign key then add foreign key after data insertion? – Vamsi Pavan Mahesh Mar 27 '15 at 12:26
@VamsiPavanMahesh, NO, cause even if you do that; your foriegn key creation will fail with same kind of error since there will be key data mismatch. – Rahul Mar 27 '15 at 18:47
8

Simply speaking if children have parent ids defined, these parents have to exist. And I thought my syntax was wrong... It helped me a lot. Thanks! – wst Apr 19 '16 at 02:11
7

What about optional relations? – Hamed Hamedi Mar 06 '19 at 16:57
@Rahul but in my case data is existed in parent table but at a sudden 3 years old database (child table) stop saving the data https://i.stack.imgur.com/dMl6m.png – Abdul Rehman Sep 17 '21 at 22:20
@HamedHamedi see Mr-Faizan's answer where he suggests this to temporarily disable any foreign key checks: SET FOREIGN_KEY_CHECKS=0; (but be sure to run this afterwards to re-enable them: SET FOREIGN_KEY_CHECKS=1) – jbobbins Oct 26 '21 at 23:30
Also have a look at my question https://stackoverflow.com/questions/76840478/sqliteconstraintexception-foreign-key-constraint-failed-code-787 – Taimoor Khan Aug 05 '23 at 07:35

score 107 · Answer 2 · answered Mar 27 '19 at 08:18

107

The Problem is with FOREIGN KEY Constraint. By Default (SET FOREIGN_KEY_CHECKS = 1). FOREIGN_KEY_CHECKS option specifies whether or not to check foreign key constraints for InnoDB tables. MySQL - SET FOREIGN_KEY_CHECKS

We can set foreign key check as disable before running Query. Disable Foreign key.

Execute one of these lines before running your query, then you can run your query successfully. :)

1) For Session (recommended)

SET FOREIGN_KEY_CHECKS=0;

2) Globally

SET GLOBAL FOREIGN_KEY_CHECKS=0;

answered Mar 27 '19 at 08:18

Mr-Faizan

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Great! It fixed my issue. – sudhee_bsp Nov 11 '20 at 08:49
4

Yeah but the important thing is, after setting the foreign key constraint to 0, don't forget to set it back – Klaus M. Sep 09 '21 at 09:33
This is basically adding a foreign key constrain to guarantee data integrity (making sure the "Ordre" that is referenced exists) and then disable checking for it while inserting the reference. – dehart Mar 17 '22 at 09:00

Madhur Bhaiya · Answer 3 · 2020-07-22T03:13:01.970

This error generally occurs because we have some values in the referencing field of the child table, which do not exist in the referenced/candidate field of the parent table.

Sometimes, we may get this error when we are applying Foreign Key constraints to existing table(s), having data in them already. Some of the other answers are suggesting to delete the data completely from child table, and then apply the constraint. However, this is not an option when we already have working/production data in the child table. In most scenarios, we will need to update the data in the child table (instead of deleting them).

Now, we can utilize Left Join to find all those rows in the child table, which does not have matching values in the parent table. Following query would be helpful to fetch those non-matching rows:

SELECT child_table.* 
FROM child_table 
LEFT JOIN parent_table 
  ON parent_table.referenced_column = child_table.referencing_column 
WHERE parent_table.referenced_column IS NULL

Now, you can generally do one (or more) of the following steps to fix the data.

Based on your "business logic", you will need to update/match these unmatching value(s), with the existing values in the parent table. You may sometimes need to set them null as well.
Delete these rows having unmatching values.
Add new rows in your parent table, corresponding to the unmatching values in the child table.

Once the data is fixed, we can apply the Foreign key constraint using ALTER TABLE syntax.

I delete all data from table and then add foreign key it has been added thank you — Sumit Kumar Gupta, Dec 29 '19 at 09:17

score 48 · Answer 4 · edited Oct 25 '19 at 16:26

48

You are getting this constraint check because Ordre table does not have reference OrdreID provided in insert command.

To insert value in Ordrelinje, you first have to enter value in Ordre table and use same OrdreID in Orderlinje table.

Or you can remove not null constraint and insert a NULL value in it.

edited Oct 25 '19 at 16:26

Aleksi Baklanov

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answered Feb 09 '14 at 14:32

Ankit Bansal

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score 47 · Answer 5 · edited Jan 23 '23 at 08:25

47

You must delete data in the child table which does not have any corresponding foreign key value to the parent table primary key .Or delete all data from the child table then insert new data having the same foreign key value as the primary key in the parent table . That should work .

edited Jan 23 '23 at 08:25

Community

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answered Mar 06 '18 at 07:24

Humphrey

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This is correct, before inserting data into the pivot table (that having CASCADE constraints), you must insert data into parent tables. e.g 1st table - permissions; 2nd table - roles; 3rd table - permission_role; So 1st Insert into permissions table, 2nd insert into roles table and at last insert into permission_role table. – Deepak Panwar Jan 07 '20 at 07:47

score 16 · Answer 6 · answered Jan 16 '20 at 11:40

16

This helped me out after reading @Mr-Faizan's and other answers.

Untick the 'Enable foreign key checks'

in phpMyAdmin and hit the query. I don't know about WorkBench but the other answers might help you out.

answered Jan 16 '20 at 11:40

Sagar Wankhede

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this is handy and worked. I had an existing DB so I had to manage with existing data. – Robert Apr 03 '23 at 18:20

score 6 · Answer 7 · answered Sep 23 '20 at 12:11

I had the same problem. I was creating relationships on existing tables but had different column values, which were supposed/assumed to be related. For example, I had a table USERS that had a column USERID with rows 1,2,3,4,5. Then I had another child table ORDERS with a column USERID with rows 1,2,3,4,5,6,7. Then I run MySQl command ALTER TABLE ORDERS ADD CONSTRAINT ORDER_TO_USER_CONS FOREIGN KEY (ORDERUSERID) REFERENCES USERS(USERID) ON DELETE SET NULL ON UPDATE CASCADE;

It was rejected with the message:

Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails (DBNAME1.#sql-4c73_c0, CONSTRAINT ORDER_TO_USER_CONS FOREIGN KEY (ORDERUSERID) REFERENCES USERS (USERID) ON DELETE SET NULL ON UPDATE CASCADE)

I exported data from the ORDERS table, then deleted all data from it, re-run the command again, it worked this time, then re-inserted the data with the corresponding USERIDs from the USERS table.

Spinstaz · Answer 8 · 2021-11-12T22:20:02.030

6

I found that changing the foreign key back from not null to null BEFORE I tried to do what I knew was the correct code, got it working. Helped that I was using Mysql workbench. I had to also set SET FOREIGN_KEY_CHECKS=0; and then back to =1; after finished.

edited Nov 12 '21 at 22:20

answered Nov 12 '21 at 22:01

Spinstaz

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score 5 · Answer 9 · answered Jul 26 '19 at 05:13

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in the foreign key table has a value that is not owned in the primary key table that will be related, so you must delete all data first / adjust the value of your foreign key table according to the value that is in your primary key

answered Jul 26 '19 at 05:13

Arbahud Rio Daroyni

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score 3 · Answer 10 · edited May 16 '19 at 14:33

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This can be fixed by inserting the respective records in the Parent table first and then we can insert records in the Child table's respective column. Also check the data type and size of the column. It should be same as the parent table column,even the engine and collation should also be the same. TRY THIS! This is how I solved mine. Correct me if am wrong.

edited May 16 '19 at 14:33

RamenChef

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answered May 16 '19 at 10:52

frz

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Edoardo Vignati · Answer 11 · 2020-09-09T08:13:32.820

In my case the tables were perfectly consistent.

Anyway I was getting this error because I created (by accident) more than one FK constraint on the same field.

I run the following query to show all the keys:

SELECT *
FROM information_schema.table_constraints
WHERE constraint_schema = 'my_db_name'

and I deleted the wrong ones with the following query:

ALTER TABLE my_table
DROP FOREIGN KEY wrong_fk_constraint;

You can check it also running this query:

SHOW CREATE TABLE my_table;

score 2 · Answer 12 · edited Oct 29 '16 at 11:24

2

While inserting the foreign key attribute values, first verify the attributes type, as well as primary key attribute value in the parent relation, if the values in parent relation matches, then you can easily insert/update child attribute values.

edited Oct 29 '16 at 11:24

Ahmad

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answered Oct 24 '16 at 13:30

ashok

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score 2 · Answer 13 · answered Dec 17 '19 at 16:47

I was getting this issue even though my parent table had all the values I was referencing in my child table. The issue seemed to be that I could not add multiple child references to a single foreign key. In other words if I had five rows of data referenced the same foreign key, MySQL was only allowing me to upload the first row and giving me the error 1452.

What worked for me was typing the code "SET GLOBAL FOREIGN_KEY_CHECKS=0". After that I closed out of MySQL and then restarted it and I was able to upload all of my data with no errors. I then typed "SET GLOBAL FOREIGN_KEY_CHECKS=1" to set the system back to normal although I'm not entirely sure what FOREIGN_KEY_CHECKS does. Hope this helps!

Isn't this just what has already been suggested in the answer posted by @Mr-Faizan? If not, then please explain what your answer adds. — Adrian Mole, Dec 17 '19 at 17:08

score 2 · Answer 14 · edited Feb 14 '20 at 12:54

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The problem occurs because you set the foreign key in child table after you insert some data in the child table.

Try removing all the data from child table, then set the foreign key and afterwards add/insert the data in table, it will work.

edited Feb 14 '20 at 12:54

THess

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answered Feb 14 '20 at 12:11

Khalid Teli

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score 2 · Answer 15 · answered Aug 18 '20 at 11:44

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First allow NULL on the parent table and set the default values to NULL. Next create the foreign key relationship. Afterwards, you can update the values to match accordingly

answered Aug 18 '20 at 11:44

Jeff Ogah

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score 2 · Answer 16 · answered Aug 06 '21 at 17:33

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For PhpMyAdmin , Go to the structure of table where you created foreign key and then click on Relation view , in that choose No Action under on Delete and on Update section. Note : this work for me. enter image description here

answered Aug 06 '21 at 17:33

vrushant domde

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score 1 · Answer 17 · answered Jul 08 '16 at 23:23

Your ORDRELINJE table is linked with ORDER table using a foreign key constraint constraint Ordrelinje_fk foreign key(Ordre) references Ordre(OrdreID) according to which Ordre int NOT NULL, column of table ORDRELINJE must match any Ordre int NOT NULL, column of ORDER table.

Now what is happening here is, when you are inserting new row into ORDRELINJE table, according to fk constraint Ordrelinje_fk it is checking ORDER table if the OrdreID is present or not and as it is not matching with any OrderId, Compiler is complaining for foreign key violation. This is the reason you are getting this error.

Foreign key is the primary key of the other table which you use in any table to link between both. This key is bound by the foreign key constraint which you specify while creating the table. Any operation on the data must not violate this constraint. Violation of this constraint may result in errors like this.

Hope I made it clear.

score 1 · Answer 18 · answered Aug 27 '18 at 20:12

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you should insert at least one raw in each tables (the ones you want the foreign keys pointing at) then you can insert or update the values of the foreign keys

answered Aug 27 '18 at 20:12

Harvester Haidar

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mamal · Answer 19 · 2019-04-29T04:12:56.970

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you should add data from REFERENCES KEY in PRIMARY TABLE to FOREIGN KEY in CHILD TABLE
it means do not add random data to foreign key ، just use data from primary key that is accessable

description of data in foreign key

edited Apr 29 '19 at 04:12

answered Dec 31 '18 at 10:31

mamal

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score 1 · Answer 20 · answered Jun 11 '20 at 05:50

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check the no. of record in parent table that matches with child table and also primary key must match with foreign key reference. This works for me.

answered Jun 11 '20 at 05:50

UC57

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score 1 · Answer 21 · answered Jan 07 '21 at 16:09

I am having the same issue here is my scenario

i put empty('') where value is NULL now this '' value does not match with the parent table's id

here is things need to check , all value with presented in parent table otherwise remove data from parent table then try

score 0 · Answer 22 · answered Nov 14 '19 at 09:21

ill squeeze this in here: my case was trying to create a like for a post which dint exist; while committing to database the error was raised. solution was to first create the post then like it. from my understanding if the post_id was to be saved in the likes table it had to first check with posts table to ascertain existence. i found it better to have it this way since its more logical to me that way..

score 0 · Answer 23 · answered Dec 11 '19 at 05:31

When you're using foreign key, your order of columns should be same for insertion.

For example, if you're adding (userid, password) in table1 from table2 then from table2 order should be same (userid, password) and not like (password,userid) where userid is foreign key in table2 of table1.

score 0 · Answer 24 · edited Dec 11 '20 at 17:17

0

The answer of your question is that you must set the same value in Primary and secondary key. Thanks

edited Dec 11 '20 at 17:17

Pirate

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answered Dec 11 '20 at 15:42

ammar

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score 0 · Answer 25 · answered Sep 17 '21 at 18:36

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Actually, i solved just like this "insert into databasename.tablename" it worked. And after when i try to add data like "insert into databasename" it worked to.

answered Sep 17 '21 at 18:36

Fazlı Keleş

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score 0 · Answer 26 · answered Nov 29 '21 at 00:21

Just something else to look for. If you had to delete records from one of your tables and are expecting the values to start at 1, you could get this error. The solution was to run a SHOW * FROM tablename on all the Parent tables. When I did I noticed in one of the tables where I had had a problem earlier and had to delete some records that the primary key values were not what I was expecting them to be from a previous SELECT *.

score 0 · Answer 27 · answered Mar 15 '22 at 19:58

Probably better answered above, but when working in mysql workbench you don't need to commit the transaction immediatly, you can commit the parent and child element at the same time. So setup the parent with sql or in the gui and add the child in the gui / sql and commit concurrently.

If working in code and getting this issue you can create a factory to create the parent and then create the child / join.

Theoretically you would need an Order to have an OrderId thus create an Order. The create an OrderId and that OrderId may have a number of associated products which you can then add to the OrderId or do with as you wish.

score 0 · Answer 28 · answered Feb 23 '23 at 14:30

Don't screw the constrains and your db structure. That is easy, check both of the columns you would like to connect again. They must have the same datatype and length. If they are not just modify the incorrect one by alter the table. If they are the same, check the data you already have within them. The logic of this relation is that you must be able to find any value from the foreign key column within one table inside the primary column from other table. As an example, let say you have:

-------------------                       -------------------
|     Table A     |                       |     Table B     |
-------------------                       -------------------
| PK | ref_id | 1 |                       | PK | b_id   | 1 |
-------------------                       -------------------
| FK | b_id   | 0 |         
-------------------

In this example, b_id column of table A could not connect to b_id column of table B because there is no record for 0 To solve that add / remove / edit the content of the tables and not the constrains.

Have fun

score 0 · Answer 29 · answered Apr 10 '23 at 16:29

I faced the same issue while using Sequelize with Nodejs/ExpressJS server and trying to insert data into a News table which had a foreign key to a Users table, basically, the user who created the news.

In my case, the problem was I had created a foreign key reference from the News table to Users table using .belongsTo() method provided by the sequelize API. Though I had not referenced a foreign key from Users to News using .hasMany(). Once I did this, then re-populated my table with data using Model.create({}), it worked!

ERROR 1452: Cannot add or update a child row: a foreign key constraint fails

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