Foreign Key Constraint Incorrectly Formed - mysql

Question

I'm having an issue in which I can't create some tables because their foreign keys are "incorrectly formed." I don't know why I'm getting this error because the keys are of the same type and have unique names. The code is pretty straightforward:
CREATE TABLE Vehicle(
vin VARCHAR(25),
ID VARCHAR(20),
make VARCHAR(20),
model VARCHAR(20),
year_ VARCHAR(20),
condition_ VARCHAR(20),
PRIMARY KEY(vin, ID),
FOREIGN KEY(ID) REFERENCES Listing(CarID)
);
CREATE TABLE Listing(
CarID VARCHAR(20),
state_ VARCHAR(20),
price INT(10),
url VARCHAR(50),
PRIMARY KEY(CarID),
FOREIGN KEY(CarID) REFERENCES Vehicle(ID)
);

Answer 1

If the primary key of Vehicle is (vin, ID) then a foreign key that references it must also have two columns, in the same order, with the same data types.
But in your case, your Listing table doesn't have a vin column, so it can't reference the primary key of Vehicle.
What would it mean for Listing to reference part of the primary key of Vehicle? There is no guarantee that only one row exists with a given value of ID in Vehicle. It could have multiple rows with the same ID and different vin values. So a Listing could potentially be the child of multiple vehicles, which probably doesn't make sense.
So you must either:
Add a vin column to Listing, so its foreign key can reference exactly one row by its primary key (both columns).
Modify the Vehicle table to make ID alone be its primary key.
Update:
I just noticed that you appear to have foreign keys in both tables. That's not usually needed. As I think about what you are trying to model, I would guess that a Listing has one or more Vehicles, right? So the Vehicle should reference its parent Listing. But a Listing does not need to reference a Vehicle.
So the following works:
First, create the parent table, because you can't make a foreign key until the parent table exists. In your case, the Listing is the parent table.
CREATE TABLE Listing(
CarID VARCHAR(20),
state_ VARCHAR(20),
price INT(10),
url VARCHAR(50),
PRIMARY KEY(CarID)
);
Then create the child table, with a foreign key that references its parent.
CREATE TABLE Vehicle(
vin VARCHAR(25),
ID VARCHAR(20),
make VARCHAR(20),
model VARCHAR(20),
year_ VARCHAR(20),
condition_ VARCHAR(20),
PRIMARY KEY(vin, ID),
FOREIGN KEY(ID) REFERENCES Listing(CarID)
);
I tested on MySQL 5.7.27.

Answer 2

This is the correct way:
CREATE TABLE Vehicle(
vin VARCHAR(25),
ID VARCHAR(20),
make VARCHAR(20),
model VARCHAR(20),
year_ VARCHAR(20),
condition_ VARCHAR(20),
PRIMARY KEY(vin, ID)
);
CREATE TABLE Listing(
CarVin VARCHAR(25),
CarID VARCHAR(20),
state_ VARCHAR(20),
price INT(10),
url VARCHAR(50),
PRIMARY KEY(CarVin, CarID),
FOREIGN KEY(CarVin, CarID) REFERENCES Vehicle(vin, ID)
);
In your data model there are a listing and vehicles.
A vehicle can be in the Listing (or not). The Listing have n vehicles (so you have a 1 to n relation).
Each row in Listing have a reference to a vehicle, so it need the vehicle primary key fields (or some unique fields) to reference to it (The vehicle primary key is compound by "vin" and "ID" fields).
So the primary key in Listing should be compound by "vin" and "ID" too, like in Vehicle's table. It's a good practice to use integer autoincrement field as primary key for each table.
You need to add in table Listing "CarVin" field and then the foreign key in Listing will be FOREIGN KEY(CarVin, CarID) REFERENCES Vehicle(vin, ID).
The foreign key will check, when you insert a Listing row, that the vehicle with the values CarVin and CarID exist.
To add a foreign key in table Vehicle too, you should disable foreign key check (SET FOREIGN_KEY_CHECKS=0;) until the tables will be created, because table Listing doesn't exist yet (if you create first table Vehicle). But I think it doesn't have sense in this case, at least you want ensure that all Vehicles are in Listing table.
To insert data you have to insert first a vehicle, and then add it to Listing table (foreign key is check, at least you disable foreign key check until the end), otherwise the foreign key check will fail.
In #"Bill Karwin"'s model you need first insert the Listing row and then the Vehicle.
MySql foreign key constraint documentation, each storage engine in MySql implements foreign key constraint in different way.
Both tables with foreign key constraint to each other:
SET FOREIGN_KEY_CHECKS=0;
CREATE TABLE Vehicle(
vin VARCHAR(25),
ID VARCHAR(20),
make VARCHAR(20),
model VARCHAR(20),
year_ VARCHAR(20),
condition_ VARCHAR(20),
PRIMARY KEY(vin, ID),
FOREIGN KEY(vin, ID) REFERENCES Listing(CarVin, CarID)
);
CREATE TABLE Listing(
CarVin VARCHAR(25),
CarID VARCHAR(20),
state_ VARCHAR(20),
price INT(10),
url VARCHAR(50),
PRIMARY KEY(CarVin, CarID),
FOREIGN KEY(CarVin, CarID) REFERENCES Vehicle(vin, ID)
);
SET FOREIGN_KEY_CHECKS = 1;