foreign-keys

Is there a easy way to query a table for its constraints(foreignkeys specificaly) like show create table, but for the constraints only? thanks, pvgoddijn To show only the foreign key constraints you can check the constraint_type in information_schema.table_constraints and get the affected columns in information_schema.key_column_usage via a join SELECT b.table_name, b.column_name, b.constraint_name, b.referenced_table_name, b.referenced_column_name FROM information_schema.table_constraints a JOIN information_schema.key_column_usage b ON a.table_schema = b.table_schema AND a.constraint_name = b

This one is for the Grails users here. I asked it on the grails - user mailing list, but I figured since I've been fighting this for a couple of days I should cast as wide a net as possible. I'm having some difficulty with trying to model relationships between two objects of the same type in another object (different type) referencing the two objects. As an example of what I'm trying to do, suppose you're modeling relationships among family members. Any given relationship "belongsTo" two different family members. So: class Person { hasMany[relationships: Relationship] static mappedBy =

I have this table: CREATE TABLE IF NOT EXISTS `produtos` ( `id` int(11) NOT NULL auto_increment, `idcatprodutos` int(11) NOT NULL, `idcategoria` int(11) NOT NULL, `idmarca` int(11) NOT NULL, `nome` varchar(100) NOT NULL, PRIMARY KEY (`id`), KEY `FK_produtos_2` (`idcatprodutos`), KEY `FK_produtos_3` (`idmarca`), KEY `FK_produtos_4` (`idcategoria`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 ROW_FORMAT=DYNAMIC AUTO_INCREMENT=39 ; and this table: CREATE TABLE IF NOT EXISTS `sugestoes` ( `id` int(11) NOT NULL auto_increment, `idproduto` int(11) NOT NULL, `idsugestao1` int(11) NOT NULL, `idsugestao2`

I'm in the process of creating a social network. It has several entities like news, photo, which can have comments. Since all comments have the same columns and behave the same way, and the only difference is their type — news, or photo, or something else to be added in the future — I decided to create one table for all comments with a column named type . It worked perfectly until I decided to add foreign keys to my database schema. The comment table have a column parent , which refers to id of news or photo table, depending on the column type . The problem is, I can't add a foreign key which

I know that this question has already been asked, but I still can't find what I'm doing wrong. I'm using the framework Laravel. I have 2 tables (Users and Locations). When I want to create a User, I get the error message: SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails ( festival_aid . users , CONSTRAINT fk_users_locations1 FOREIGN KEY ( location_id ) REFERENCES locations ( location_id ) ON DELETE CASCADE ON UPDATE NO ACTION) (SQL: insert into users ( user_id , user_email , location_id ) values (?, ?, ?)) (Bindings: array (

SQLite v3.7.5 Is there a way to enable SQLite Foreign Keys with cascade delete enabled by default? Given the following example: CREATE TABLE [Parent] ( [ParentId] INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, [Name] VARCHAR(50) UNIQUE NOT NULL ); CREATE TABLE [Child] ( [ChildId] INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, [ParentId] INTEGER NOT NULL, [Name] VARCHAR(50) NOT NULL, FOREIGN KEY(ChildId) REFERENCES Child(ParentId) ON DELETE CASCADE ); The only way I've been able to enable the cascade delete is to execute the PRAGMA foreign_keys = true command before a transaction: using( var conn =

I'm building a web app that tracks what library books a person checks out. I have the following models: class Person(models.Model): name = models.CharField(max_length=100) def __unicode__(self): return self.name class Book(models.Model): name = models.CharField(max_length=100) person = models.ForeignKey(Person) checkout_date = models.DateTimeField('checkout date') def __unicode__(self): return self.name On the Admin's "change list" page for Person, I would like to show what books that person has. Is this something that can be done? If so, how? admin.py class BookAdmin(admin.ModelAdmin): list

Since foreign keys are not supported by partitioned mySQL databases for the moment, I would like to hear some pro's and con's for a read-heavy application that will handle around 1-400 000 rows per table. Unfortunately, I dont have enough experience yet in this area to make the conclusion by myself... Thanks a lot! References: How to handle foreign key while partitioning Partitioning mySQL tables that has foreign keys? Well, if you need partitioning for a table as small as 400.000 rows get another database than MySQL. Seriously. By modern standards any table below 1.000.000 rows is normally