Ошибка 1826 mysql

Normally I’d assume this is the fault of Microsoft Access, since Access frequently fails to conform to standard SQL. But this time it’s the fault of MySQL.

This is a bug in MySQL: https://bugs.mysql.com/bug.php?id=72751

Constraint names are supposed to be case-sensitive, per https://dev.mysql.com/doc/refman/8.0/en/identifier-case-sensitivity.html:

Object names may be considered duplicates if their uppercase forms are equal according to a binary collation.

In other words, some object names are case-insensitive.

That is true for names of cursors, conditions, procedures, functions, savepoints, stored routine parameters, stored program local variables, and plugins. It is not true for names of columns, constraints, databases, partitions, statements prepared with PREPARE, tables, triggers, users, and user-defined variables.

That says that constraint names are one of the object types whose names are supposed to case-sensitive.

But that documentation is incorrect. Currently MySQL treats constraint names as case-insensitive. So constraints named idMarca and IdMarca conflict.

We can demo this:

mysql> create table parent (id int primary key);
Query OK, 0 rows affected (0.01 sec)

mysql> create table child1 (parent int, constraint con foreign key (parent) references parent(id));
Query OK, 0 rows affected (0.01 sec)

mysql> create table child2 (parent int, constraint Con foreign key (parent) references parent(id));
ERROR 1826 (HY000): Duplicate foreign key constraint name 'Con'

That’s the error message in MySQL 8.0. The names con and Con should be treated as distinct, but they aren’t.

In older versions of MySQL, the error message wasn’t clear, something like «1050: Table ‘./test/child2’ already exists».

How to fix this? You’ll have to create your foreign key constraints manually.

This bug was reported in 2014, and hasn’t been fixed, so I wouldn’t get your hopes up that it will be fixed soon. You may click the «Affects Me» button in the bug tracker to vote for it to get some attention, but don’t count on it. Just fix the constraint names in your project.

Я пишу код на MySQL, и это дает мне ошибки.

Это мой код, и он выдает ошибку — «Код ошибки: 1826. Повторяющееся имя ограничения внешнего ключа ‘menu_ibfk_1’»

Когда я даю другое имя ограничения, появляется ошибка — «Код ошибки: 1822. Не удалось добавить ограничение внешнего ключа. Отсутствует индекс для ограничения ‘menu_ibfk_2’ в указанной таблице ‘menu’»

create database WasteManagement;
CREATE TABLE WasteManagement.Faculty_login ( 
Faculty_ID int(9) NOT NULL  ,
FName varchar(50) DEFAULT NULL,
Department varchar(20) DEFAULT NULL, 
Password varchar(20) DEFAULT NULL, 
PRIMARY KEY (Faculty_ID) );

CREATE TABLE WasteManagement.Student_login ( 
Enrollment_No int(11) NOT NULL , 
SName varchar(50) DEFAULT NULL, 
Course varchar(20) DEFAULT NULL, 
Password varchar(20) DEFAULT NULL, 
PRIMARY KEY (Enrollment_No) );

CREATE TABLE WasteManagement.Staff_login ( 
Staff_ID int(9) NOT NULL  ,
CName varchar(50) DEFAULT NULL,
Username varchar(20) DEFAULT NULL, 
Password varchar(20) DEFAULT NULL, 
PRIMARY KEY (Staff_ID) );

CREATE TABLE WasteManagement.Menu ( 
Staff_ID int(9) NOT NULL,
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
PRIMARY KEY (Datee, Timee, Dish)
#FOREIGN KEY (Staff_ID) References Menu(Staff_ID) 
);

CREATE TABLE WasteManagement.Demand1 ( 
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
S_demand BOOLEAN DEFAULT FALSE, 
PRIMARY KEY (Datee, Timee, Dish, S_demand),
KEY d1 (datee),
KEY t1 (Timee),
KEY dish1 (Dish),
#KEY `AuthorID` (`AuthorID`),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Datee) References Menu(Datee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Timee) References Menu(Timee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Dish) References Menu(Dish));

CREATE TABLE WasteManagement.Demand2 ( 
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
F_demand BOOLEAN NOT NULL DEFAULT FALSE, 
PRIMARY KEY (Datee, Timee, Dish, F_demand),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Datee) References Menu(Datee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Timee) References Menu(Timee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Dish) References Menu(Dish));

CREATE TABLE WasteManagement.Feedback1 ( 
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
eaten_by1 int(7) NOT NULL, 
s_rating int(1) NOT NULL,
PRIMARY KEY (Datee, Timee, Dish, eaten_by1, s_rating),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Datee) References Menu(Datee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Timee) References Menu(Timee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Dish) References Menu(Dish));

CREATE TABLE WasteManagement.Feedback2 ( 
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
eaten_by2 int(7) NOT NULL, 
f_rating int(1) NOT NULL,
PRIMARY KEY (Datee, Timee, Dish, eaten_by2, f_rating),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Datee) References Menu(Datee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Timee) References Menu(Timee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Dish) References Menu (Dish));

CREATE TABLE WasteManagement.Wastage ( 
Datee date NOT NULL DEFAULT 0,
Timee varchar(10) NOT NULL,
Dish varchar(30) NOT NULL, 
Produced_qty int(7) NOT NULL, 
Wasted_qty int(7) NOT NULL,
PRIMARY KEY (Datee, Time, Dish, Produced_qty, Wasted_qty),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Datee) References Menu(Datee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Timee) References Menu(Timee),
CONSTRAINT `menu_ibfk_1` FOREIGN KEY (Dish) References Menu(Dish));

Это снимок вывода MySQL

Снимок вывода MySQL

Что я делаю не так?