MariaDB - Foreign key constraint incorrect? - mariadb

Question

I tried to re-engineer a database, that I use at work. The one at work is MS Access. At home, it's MariaDB. For convenience, I use MySQL Workbench.
When sending the complete SQL dump to the server, I get an error concerning some foreign key not being correctly formed. I guess, it is a minor mistake, but still I cannot find it.
My InnoDB status tells me this:
LATEST FOREIGN KEY ERROR
2018-10-03 00:18:29 409c7450 Error in foreign key constraint of table `mydb`.`IF`:
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblBelegPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_tblBelege_tblECKassenschnittPositionen10`
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblECKassenschnittPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB:
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.
Note that the internal storage type of ENUM and SET changed in
tables created with >= InnoDB-4.1.12, and such columns in old tables
cannot be referenced by such columns in new tables.
See http://dev.mysql.com/doc/refman/5.6/en/innodb-foreign-key-constraints.html
for correct foreign key definition.
Create table '`mydb`.`IF`' with foreign key constraint failed. There is no index in the referenced table where the referenced columns appear as the first columns near '
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblBelegPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_tblBelege_tblECKassenschnittPositionen10`
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblECKassenschnittPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB'.
The really weird thing is that I do not have any table named "IF"...
Can anyone make heads or tails of this for me? That would be very much appreciated.
-- Table `mydb`.`tblBelegPositionen`
-- -----------------------------------------------------
DROP TABLE IF EXISTS `mydb`.`tblBelegPositionen` ;
SHOW WARNINGS;
CREATE TABLE IF NOT EXISTS `mydb`.`tblBelegPositionen` (
`belegposid` INT NOT NULL AUTO_INCREMENT,
`belegposBetrag` DOUBLE NOT NULL,
`zahlartfID` INT NOT NULL,
`belegfID` INT NOT NULL,
PRIMARY KEY (`belegposid`))
ENGINE = InnoDB;
SHOW WARNINGS;
-- Table `mydb`.`tblECKassenschnittPositionen`
-- -----------------------------------------------------
DROP TABLE IF EXISTS `mydb`.`tblECKassenschnittPositionen` ;
SHOW WARNINGS;
CREATE TABLE IF NOT EXISTS `mydb`.`tblECKassenschnittPositionen` (
`ecposid` INT NOT NULL AUTO_INCREMENT,
`belegfID` INT NOT NULL,
`ecposBetrag` DOUBLE NOT NULL,
`kassenschnittfID` INT NOT NULL,
PRIMARY KEY (`ecposid`))
ENGINE = InnoDB;
SHOW WARNINGS;
-- Table `mydb`.`tblBelege`
-- -----------------------------------------------------
DROP TABLE IF EXISTS `mydb`.`tblBelege` ;
SHOW WARNINGS;
CREATE TABLE IF NOT EXISTS `mydb`.`tblBelege` (
`belegid` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`belegKassierer` INT NOT NULL,
`belegDatum` DATETIME NOT NULL,
`kassefID` INT NOT NULL,
`belegSchicht` INT NULL,
`gvfID` INT NOT NULL,
`belegJahr` YEAR NULL,
`belegDruckErfolgt` TINYINT(1) NULL,
`belegDruckDatum` DATETIME NULL,
`belegPeriodenfremdeBuchung` TINYINT(1) NULL,
PRIMARY KEY (`belegid`, `gvfID`, `kassefID`),
CONSTRAINT `fk_tblBelege_tblBelegPositionen10`
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblBelegPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_tblBelege_tblECKassenschnittPositionen10`
FOREIGN KEY (`belegid`)
REFERENCES `mydb`.`tblECKassenschnittPositionen` (`belegfID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
SHOW WARNINGS;

Answer 1

Ok, so there are a couple of things to know here, as there were a couple of errors. You must check all the items that the error mentions, for correctness, to avoid trouble.
The cannot find an index in the referenced table portion of the error message means that the column/field specified in the REFERENCES clause must be indexed in the other table.
Then, the column type definition of the column specified in the FOREIGN KEY clause must match the column type of the column specified in the REFERENCES clause too, so even though the first item corrects part of the problem, there will still be an error related to another portion of the message: ...or column types in the table and the referenced table do not match.
So, to fix item 1, run these 2 queries:
ALTER TABLE `tblbelegpositionen` ADD INDEX(`belegfID`);
ALTER TABLE `tbleckassenschnittpositionen` ADD INDEX(`belegfID`);
Then to fix item 2, I had to change the first column of table tblBelege
from this:
`belegid` INT UNSIGNED NOT NULL AUTO_INCREMENT,
to this:
`belegid` INT NOT NULL,
...so that they matched the same type as the belegfID column as defined in the other tables. After those two changes, I was able to successfully run your CREATE TABLE `tblBelege` statement.
So, to recap:
Run your create table statements for tblbelegpositionen and
tbleckassenschnittpositionen.
Then run the 2 ALTER statements shown above for item 1.
Modify the first column of table tblBelege to match the column types as defined to belegfID in the other tables for item 2.
Then run your modified CREATE TABLE statement (with the item 2 change applied) to create the tblBelege table.
I'm a little confused about the same FOREIGN KEY referencing 2 different tables, but if it works for you, then ok. (not saying that it cannot be done, I've never used a foreign key that way) Perhaps you meant the opposite, to have a foreign key in the other 2 tables (1 in each table) that refer to tblBelege instead? If so, then you could add unsigned to the type definition for belegfID and it would work, and would not need the change that I mentioned with item 2.
Oh, and after you run the ALTER statements, you can view the table structure by running:
SHOW CREATE TABLE `tblbelegpositionen`;
SHOW CREATE TABLE `tbleckassenschnittpositionen`;
...to get the KEY definition that was added to include with your CREATE TABLE statements. Since they both create the same key, you really only need to run one of those statements and then add the KEY definition to both table statements.