I’m trying to add barista_grade_id to the baristas table as a foreign key and I’ve looked every where on what syntax to use and it seems to be unanimous that this is the correct way. I did it before this and it worked. but because of some mistakes I deleted it and redo it again, but for whatever reason it
Tag: foreign-keys
Insert with Foreign Key Contraints
In the above code, I want to add the distinct values from column make from vehicles into column makeID in makeModel but I get the error INSERT INTO makeModel (makeID) SELECT DISTINCT (make) FROM vehicle Error Code: 1452. Cannot add or update a child row: a foreign key constraint fails (`dealership`.`makemodel`, CONSTRAINT `makemodel_ibfk_2` FOREIGN KEY (`modelID`) REFERENCES `model` (`ID`)) Answer
SQL FOREIGN KEY ERROR (errno: 150 “Foreign key constraint is incorrectly formed”)
Hi I have this three very simple tables but I can’t fix it to get the right format of foreign key. CREATE TABLE company( company_name varchar(30) UNIQUE NOT NULL, bid INT(15) NOT NULL UNIQUE, cid INT(15) NOT NULL UNIQUE, FOREIGN KEY (bid) REFERENCES branch(branch_id), FOREIGN KEY (cid) REFERENCES contact(contact_id) ); CREATE TABLE branch( branch_id INT(15) NOT NULL AUTO_INCREMENT UNIQUE, branch_type
Hasura Graphql does not return all documents when multiple foreign keys exist in a table to the same foreign field
Query: Response: Expected: Why are the last two documents not showing up from the first query? Connection table definition: I know this could be more of a SQL question, but it would be great if you can show the graphql version too Answer You have to create two relationships between users table and connections table, Then query : Output:
MYSQL table wont allow multiple foreign keys
I know this has been asked again and again, and I’ve tried so many times and don’t understand why I keep getting errors, but I’m trying to connect the order details table to the order items, users and payment table, but SQL is coming up with. (this is for a school project) I’ve been able to connect a table with
How to use constraints to force two child items be from the same parent?
I have a Jobs table that holds jobs. I have a Tasks table that holds tasks that belong to a job (1:many). I have a Task_Relationships table that holds the data about which tasks depend on other tasks within a job. I have 2 jobs, each job has 3 tasks and within the jobs the tasks are related as in
mysql error 1822 when trying to create new table with a FOREIGN KEY
trying the following code: Give back Error Code: 1822. Failed to add the foreign key constraint. Missing index for constraint ‘stock_ibfk_1’ in the referenced table ‘donation’. Why? how can I solve this problem? Answer to create a foreign key relationship, the parent table column on which you are creating relation must be unique or primary and they must have the
MYSQL – Cannot update a parent row: a foreign key constraint fails
db-fiddle link SCHEMA QUERY UPDATE QUERY ERROR If I change the ID of FIELD_TEST, To change the GRAPH_FIELD_INFO_TEST ID referring to the ID of FIELD_TEST, I set it as cascade in GRAPH_FIELD_INFO_TEST. However, it fails due to the following error. If I change the ID of GRAPH, it works normally, but it doesn’t work properly only if the ID of
How does foreign key resolve queries in terms of performance(indexing) on databases?
I’m wondering how queries with join operation perform when I have or don’t have a foreign key. For example, given the following schema: And given I have the following query: When I ask for the execution plan of it, it gives me the result below. It means the query is using the index I_EMP_DEPTID to improve the join operation. However,
Failed to add the foreign key constraint. Missing index for constraint ‘information_ibfk_1’ in the referenced table ‘informacia’
What is the problem with these tables? I want to create foreign key for second one Answer In order for a table to add a foreign key constraint, the referenced column (zip_code in this case) must be a PRIMARY KEY or must have a UNIQUE constraint (ideally with NOT NULL). I modified the SQL statements according to PostgreSQL database. You