Question

I created the master table with the composite primary key.

parent table structure is as follows:

CREATE TABLE `taskcategory` (
  `SiteID` int(10) unsigned NOT NULL DEFAULT 1,
  `TaskID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `TaskName` varchar(45) DEFAULT '',
  `TaskDescription` varchar(45) DEFAULT '',
  `IsInbuild` int(11) DEFAULT '1',
  PRIMARY KEY (`TaskID`,`SiteID`)
);

when i am trying to create the table with foreign key with the above parent table reference i am getting ‘can’t create table error no 150’ error . help me to do that.

child table structure as follows:

CREATE TABLE taskdetails (`SiteID` int(10) unsigned NOT NULL DEFAULT '1',
  `TaskID` int(10) unsigned NOT NULL DEFAULT '0',
  `SubtaskID` int(10) unsigned NOT NULL,
  `ScriptName` varchar(255) DEFAULT '',
  `FunctionName` varchar(255) DEFAULT '',
  `ButtonName` varchar(255) DEFAULT '',
  `IsInbuild` int(10) unsigned DEFAULT '1',
  `Description` varchar(255) DEFAULT '',
  PRIMARY KEY (`SubtaskID`,`TaskID`,`SiteID`),
                            INDEX (siteid, taskid),
                            FOREIGN KEY (siteid, taskid)
                              REFERENCES taskcategory(siteid, taskid)
                              ON UPDATE CASCADE ON DELETE CASCADE
                            ) ENGINE=INNODB;

help me to resolve it.

Answer 1

From the manual:

InnoDB requires indexes on foreign keys and referenced keys so that
foreign key checks can be fast and not require a table scan. In the
referencing table, there must be an index where the foreign key
columns are listed as the first columns in the same order. Such an
index is created on the referencing table automatically if it does not
exist. This index might be silently dropped later, if you create
another index that can be used to enforce the foreign key constraint.
index_name, if given, is used as described previously.

So when you add an index in the parent table it works (yes, I tested it):

CREATE TABLE `taskcategory` (
  `SiteID` int(10) unsigned NOT NULL DEFAULT 1,
  `TaskID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `TaskName` varchar(45) DEFAULT '',
  `TaskDescription` varchar(45) DEFAULT '',
  `IsInbuild` int(11) DEFAULT '1',
  PRIMARY KEY (`TaskID`,`SiteID`)
  , INDEX (SiteID, TaskID)
) ENGINE=INNODB;

CREATE TABLE taskdetails (`SiteID` int(10) unsigned NOT NULL DEFAULT '1',
  `TaskID` int(10) unsigned NOT NULL DEFAULT '0',
  `SubtaskID` int(10) unsigned NOT NULL,
  `ScriptName` varchar(255) DEFAULT '',
  `FunctionName` varchar(255) DEFAULT '',
  `ButtonName` varchar(255) DEFAULT '',
  `IsInbuild` int(10) unsigned DEFAULT '1',
  `Description` varchar(255) DEFAULT '',
  PRIMARY KEY (`SubtaskID`,`TaskID`,`SiteID`)
                            ,INDEX (SiteID, TaskID)
                            ,FOREIGN KEY (SiteID, TaskID)
                              REFERENCES taskcategory(SiteID, TaskID)
                              ON UPDATE CASCADE ON DELETE CASCADE
                            ) ENGINE=INNODB;

You have a primary key on those columns already (which means there’s an implicit index), but the order of the columns is important!