CODEWITHSUNDEEP
mysqlindexingforeign-keys
numfar
numfar

Reputation: 1675

MySQL cannot create foreign key constraint

I'm having some problems creating a foreign key to an existing table in a MySQL database.

I have the table exp:

+-------------+------------------+------+-----+---------+-------+
| Field       | Type             | Null | Key | Default | Extra |
+-------------+------------------+------+-----+---------+-------+
| EID         | varchar(45)      | NO   | PRI | NULL    |       |
| Comment     | text             | YES  |     | NULL    |       |
| Initials    | varchar(255)     | NO   |     | NULL    |       |
| ExpDate     | date             | NO   |     | NULL    |       |
| InsertDate  | date             | NO   |     | NULL    |       |
| inserted_by | int(11) unsigned | YES  | MUL | NULL    |       |
+-------------+------------------+------+-----+---------+-------+

and I wan't to create a new table called sample_df referencing this, using the following:

CREATE TABLE sample_df (
    df_id         mediumint(5) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    sample_type   mediumint(5) UNSIGNED          NOT NULL,
    df_10         boolean                        NOT NULL,
    df_100        boolean                        NOT NULL,
    df_1000       boolean                        NOT NULL,
    df_above_1000 boolean                        NOT NULL,
    target        int(11) UNSIGNED               NOT NULL,
    assay         mediumint(5) UNSIGNED ZEROFILL NOT NULL,
    insert_date   timestamp                      NOT NULL DEFAULT CURRENT_TIMESTAMP,
    inserted_by   int(11) UNSIGNED               NOT NULL,
    initials      varchar(255),
    experiment    varchar(45),
    CONSTRAINT FOREIGN KEY (inserted_by) REFERENCES user (iduser),
    CONSTRAINT FOREIGN KEY (target) REFERENCES protein (PID),
    CONSTRAINT FOREIGN KEY (sample_type) REFERENCES sample_type (ID),
    CONSTRAINT FOREIGN KEY (assay) REFERENCES assays (AID),
    CONSTRAINT FOREIGN KEY (experiment) REFERENCES exp (EID)
);

But I get the error:

ERROR 1215 (HY000): Cannot add foreign key constraint

To get some more information, I did:

SHOW ENGINE INNODB STATUS\G

From which I got:

FOREIGN KEY (experiment) REFERENCES exp (EID) 
): 
Cannot find an index in the referenced table where the 
referenced columns appear as the first columns, or column types 
in the table and the referenced table do not match for constraint.

To me, the column types seem to match, since they are both varchar(45). (I also tried setting the experiment column to not null, but this didn't fix it). So I guess the problem must be that

Cannot find an index in the referenced table where the referenced columns appear as the first columns.

But I'm not quite sure what this means, or how to check/fix it. Does anyone have any suggestions? And what is meant by first columns?

Upvotes: 119

Views: 126056

Answers (22)

Code_Ninja
Code_Ninja

Reputation: 1867

In my case, I missed that referenced column is unsigned whereas my referencing column isnt. Very silly solution, but totally missed it.

Upvotes: 0

gigibits
gigibits

Reputation: 1

Faced a similiar issue,

My first table

CREATE TABLE customers ( 
                ID int PRIMARY KEY AUTO_INCREMENT,
                 firstname varchar(255) ,
                 middlename varchar(255) ,
                 lastname varchar(255) ,
                 email varchar(255) ,
                 country varchar(255) ,
                 state varchar(255) ,
                 address varchar(255) ,
                 zipcode varchar(255) ,
                 phone varchar(255) ,
                 dob varchar(255) ,
                 password varchar(255) ,
                 accountnumber varchar(255) ,
                 ddate TIMESTAMP
                                       ) ;

My second table

CREATE TABLE transactions ( 
            ID int PRIMARY KEY AUTO_INCREMENT,
             amount INT(255) ,
             type ENUM("Debit","Credit") ,
             from_to INT(255) ,
             description varchar(255) ,
             balance INT(255) ,
             accountnumber varchar(255) ,
             tdate TIMESTAMP,
             FOREIGN KEY(accountnumber) REFERENCES customers(accountnumber)
                                   ) ;

The second table could not be created. I solved it. By altering the first table with UNIQUE for the column name accountnumber

Alter table customers add UNIQUE(accountnumber);

Then after i tried creating the second table transactions, and it worked.

So the first table should have been something like this:

CREATE TABLE customers ( 
                ID int PRIMARY KEY AUTO_INCREMENT,
                 firstname varchar(255) ,
                 middlename varchar(255) ,
                 lastname varchar(255) ,
                 email varchar(255) ,
                 country varchar(255) ,
                 state varchar(255) ,
                 address varchar(255) ,
                 zipcode varchar(255) ,
                 phone varchar(255) ,
                 dob varchar(255) ,
                 password varchar(255) ,
                 accountnumber varchar(255) UNIQUE,
                 ddate TIMESTAMP
                                       ) ;

Upvotes: 0

Marco
Marco

Reputation: 3641

In my case I referred directly to a PRIMARY KEY and got the error shown above. After adding a "normal" index additionally to my primary key it worked:

ALTER TABLE `TableName` ADD INDEX `id` (`id`);

Now it looks like this:

enter image description here

When I try to drop the INDEX id again I get following error: (1553): Cannot drop index 'id': needed in a foreign key constraint.

EDIT: This is not a new question - I just want to show the way i solved this problem for me and what kind of problems may occur.

Upvotes: -2

Gopinathsgn
Gopinathsgn

Reputation: 101

If Data type is same, Probably error is due to different Charset and Collation, try altering column as referenced column's Character set and Collate with a query something like this,

ALTER TABLE table_name MODIFY COLUMN
 column_name VARCHAR(255) CHARACTER SET utf8 COLLATE utf8_general_ci;

and then run the query for foreign key..

(alter charset of a table and database didn't work for me on incompatible error, until I alter specific column like my above suggestion)

Upvotes: 0

live-love
live-love

Reputation: 52366

I spent hours trying to get this to work. It turned out I had an older version of Heidi, 11.0.0.5919 which was not displaying the UNSIGNED attribute in the create table statement (Heidi bug), which I had used to copy from. Couldn't see it in the table design view either.

So the original table had an UNSIGNED attribute, but my foreign key didn't. The solution was upgrading Heidi, and adding the UNSIGNED attribute in the create table .

CREATE TABLE `extension` (
    `ExtensionId` INT NOT NULL,
    `AccountId` INT NOT NULL,
    PRIMARY KEY (`ExtensionId`) USING BTREE,
    INDEX `AccountId` (`AccountId`) USING BTREE,
    CONSTRAINT `AccountId` FOREIGN KEY (`AccountId`) REFERENCES  `accounts` (`AccountId`) ON UPDATE NO ACTION ON DELETE NO ACTION,
)
COLLATE='utf8mb4_0900_ai_ci'
ENGINE=InnoDB
;

Change to:

`AccountId` INT UNSIGNED NOT NULL,

Upvotes: 1

Thai Tran
Thai Tran

Reputation: 9935

It is mostly because the old table you are referring to does not have the suitable data type / collation / engine with the new table. The way to detect the difference is dumping that old table out and see how the database collect the information to have the dump script

mysqldump -uroot -p -h127.0.0.1 your_database your_old_table > dump_table.sql

It will give you enough information for you to compare

create table your_old_table
(
    `id`  varchar(32) not null,
) Engine = InnoDB DEFAULT CHARSET=utf8mb3;

This only works if you have the permission to dump your table scheme

Upvotes: 0

Brett
Brett

Reputation: 2010

Just to throw another solution in the mix. I had on delete set to set null but the field that i was putting the foreign key on was NOT nullable so making it nullable allowed the foreign key to be created.

As others have said the following things can be an issue

Field Length - INT -> BIGINT, VARCHAR(20) -> VARCHAR(40)
Unsigned - UNSIGNED -> Signed
Mixed Collations

Upvotes: 1

Ahmad Hadizadeh
Ahmad Hadizadeh

Reputation: 1

I had the same problem with writing this piece of code in the OnModelCreating method My problem was completely solved and my tables and migrations were created without errors. Please try it

var cascadeFKs = modelBuilder.Model.GetEntityTypes()
    .SelectMany(t => t.GetForeignKeys())
    .Where(fk => !fk.IsOwnership && fk.DeleteBehavior == DeleteBehavior.Cascade);

foreach (var fk in cascadeFKs)
    fk.DeleteBehavior = DeleteBehavior.Restrict;

Upvotes: 0

Андрей Пухкало
Андрей Пухкало

Reputation: 116

In my case, it was an incompatibility with ENGINE and COLLATE, once i added ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci it worked

CREATE TABLE `some_table` (
      `id` varchar(36) NOT NULL,
      `col_id` varchar(36) NOT NULL,
      
      PRIMARY KEY (`id`),
      CONSTRAINT `FK_some_table_cols_col_id` FOREIGN KEY (`col_id`) REFERENCES `ref_table` (`id`) ON DELETE CASCADE,
      
  ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

Upvotes: 4

Alex Hellier
Alex Hellier

Reputation: 485

And just to add to this , I've had the same issue today

Both fields were int of same length etc, however, one was unsigned and this was enough to break it.

Both needed to be declared as unsigned

Upvotes: 2

Diego Herrera
Diego Herrera

Reputation: 79

For me it was just the charset and collation of the DB. I changed to utf8_unicode_ci and works

Upvotes: 7

helenatxu
helenatxu

Reputation: 127

In my case was created using integer for the id, and the referencing table was creating by default a foreign key using bigint.

This caused a big nightmare in my Rails app as the migration failed but the fields were actually created in DB, so they showed up in the DB but not in the schema of the Rails app.

Upvotes: 2

sprksh
sprksh

Reputation: 2384

In some cases, I had to make the referenced field unique on top of defining it as the primary key.

But I found that not defining it as unique doesn't create a problem in every case. I have not been able to figure out the scenarios though. Probably something to do with nullable definition.

Upvotes: 1

user11891461
user11891461

Reputation: 151

Mine was a collation issue between the referenced table and the to be created table so I had to explicitly set the collation type of the key I was referencing.

  • First I ran a query at referenced table to get its collation type
show table STATUS like '<table_name_here>';
  • Then I copied the collation type and explicitly stated employee_id's collation type at the creation query. In my case it was utf8_general_ci
CREATE TABLE dbo.sample_db
(
  id INT PRIMARY KEY AUTO_INCREMENT,
  event_id INT SIGNED NOT NULL,
  employee_id varchar(45) COLLATE utf8_general_ci NOT NULL,
  event_date_time DATETIME,
  CONSTRAINT sample_db_event_event_id_fk FOREIGN KEY (event_id) REFERENCES event (event_id),
  CONSTRAINT sample_db_employee_employee_id_fk FOREIGN KEY (employee_id) REFERENCES employee (employee_id)
);

Upvotes: 15

Paul Carlton
Paul Carlton

Reputation: 2993

I had this error as well. None of the answers pertained to me. In my case, my GUI automatically creates a table with a primary unique identifier as "unassigned". This fails when I try and create a foreign key and gives me the exact same error. My primary key needs to be assigned.

If you write the SQL itself like so id int unique auto_increment then you don't have this issue but for some reason my GUI does this instead id int unassigned unique auto_increment.

Hope this helps someone else down the road.

Upvotes: 2

Nelson
Nelson

Reputation: 2177

The exact order of the primary key also needs to match with no extra columns in between.

I had a primary key setup where the column order actually matches, but the problem was the primary key had an extra column in it that is not part of the foreign key of the referencing table

e.g.) table 2, column (a, b, c) -> table 1, column (a, b, d, c) -- THIS FAILS

I had to reorder the primary key columns so that not only they're ordered the same way, but have no extra columns in the middle:

e.g.) table 2, column (a, b, c) -> table 1, column (a, b, c, d) -- THIS SUCCEEDS

Upvotes: 2

Pratik Singhal
Pratik Singhal

Reputation: 6492

This error can also occur, if the references table and the current table don't have the same character set.

Upvotes: 75

Nick Manning
Nick Manning

Reputation: 2989

In my case, it turned out the referenced column wasn't declared primary or unique.

https://stackoverflow.com/a/18435114/1763217

Upvotes: 6

Edward
Edward

Reputation: 1883

Referencing the same column more than once in the same constraint also produces this Cannot find an index in the referenced table error, but can be difficult to spot on large tables. Split up the constraints and it will work as expected.

Upvotes: 1

Austen Hoogen
Austen Hoogen

Reputation: 2826

Just throwing this into the mix of possible causes, I ran into this when the referencing table column had the same "type" but did not have the same signing.

In my case, the referenced table colum was TINYINT UNSIGNED and my referencing table column was TINYINT SIGNED. Aligning both columns solved the issue.

Upvotes: 207

gorodezkiy
gorodezkiy

Reputation: 3419

As mentioned @Anton, this could be because of the different data type. In my case I had primary key BIGINT(20) and tried to set foreight key with INT(10)

Upvotes: 25

Anton
Anton

Reputation: 559

According to http://dev.mysql.com/doc/refman/5.5/en/create-table-foreign-keys.html

MySQL requires indexes on foreign keys and referenced keys so that foreign key checks can be fast and not require a table scan. In the referencing table, there must be an index where the foreign key columns are listed as the first columns in the same order.

InnoDB permits a foreign key to reference any index column or group of columns. However, in the referenced table, there must be an index where the referenced columns are listed as the first columns in the same order.

So if the index in referenced table is exist and it is consists from several columns, and desired column is not first, the error shall be occurred.

The cause of our error was due to violation of following rule:

Corresponding columns in the foreign key and the referenced key must have similar data types. The size and sign of integer types must be the same. The length of string types need not be the same. For nonbinary (character) string columns, the character set and collation must be the same.

Upvotes: 28

Related Questions