Can't insert : A foreign key constraint fails - php

Question

I have the following tables:
CREATE TABLE IF NOT EXISTS `location`(
`ID` int(11) NOT NULL,
`name` varchar(25) NOT NULL,
`water` varchar(25) NOT NULL,
`fodder` varchar(25) NOT NULL,
`access` varchar(25) NOT NULL,
PRIMARY KEY (`ID`)
KEY `water` (`water`)
KEY `fodder` (`fodder`)
KEY `access` (`access`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;
CREATE TABLE IF NOT EXISTS `watercondition`(
`ID` int(11) NOT NULL,
`watervalue` varchar(25) NOT NULL,
PRIMARY KEY (`watervalue`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;
CREATE TABLE IF NOT EXISTS `foddercondition`(
`ID` int(11) NOT NULL,
`foddervalue` varchar(25) NOT NULL,
PRIMARY KEY (`foddervalue`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;
CREATE TABLE IF NOT EXISTS `accesscondition`(
`ID` int(11) NOT NULL,
`accessvalue` varchar(25) NOT NULL,
PRIMARY KEY (`accessvalue`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;
And for constraints table :
ALTER TABLE `location`
ADD CONSTRAINT `location_ibfk2` FOREIGN KEY (`water`) REFERENCES `watercondition` (`watervalue`),
ADD CONSTRAINT `location_ibfk3` FOREIGN KEY (`fodder`) REFERENCES `foddercondition` (`foddervalue`),
ADD CONSTRAINT `location_ibfk4` FOREIGN KEY (`access`) REFERENCES `accesscondition` (`accessvalue`);
In my php file, i want to insert a value to all of the table like this :
$sqlwater = "INSERT INTO `watercondition` (`ID`, `watervalue`) VALUES ('".$_SESSION['loc_id']."', '$watervalue')";
$resultwater = mysqli_query($con, $sqlwater) or die (mysqli_error($con));
$sqlfodder = "INSERT INTO `foddercondition` (`ID`, `foddervalue`) VALUES ('".$_SESSION['loc_id']."', '$foddervalue')";
$resultfodder = mysqli_query($con, $sqlfodder) or die (mysqli_error($con));
$sqlaccess = "INSERT INTO `accesscondition` (`ID`, `accessvalue`) VALUES ('".$_SESSION['loc_id']."', '$accessvalue')";
$resultaccess = mysqli_query($con, $access) or die (mysqli_error($con));
$sqlloc = "INSERT INTO `location` (`ID`, `name`) VALUES ('".$_SESSION['loc_id']."', '$name')";
$resultaccess = mysqli_query($con, $access) or die (mysqli_error($con));
But when I execute the php file, I get this error :
Cannot add or update a child row: a foreign key constraint fails (mydb.location, CONSTRAINT location_ibfk2 FOREIGN KEY (water) REFERENCES watercondition (watervalue))
When I check on my db, the value from water, fodder, and access have already been inserted db, but not in my location table.

Answer 1

The insert into the location table must also include values for the water, fodder and location columns. They are columns in the location table and cannot just be ignored.
Also you were using the wrong query variable in the final query.
I guess what is happening here is that the constraints are being validated by MYSQL before it checks that you have values for all the NOT NULL fields, so you get the constraint error before the more obvious one about missing column values.
$sqlloc = "INSERT INTO `location`
(`ID`, `name`, `water`, `fodder`, `location`)
VALUES ('{$_SESSION['loc_id']}', '$name',
'$watervalue', '$foddervalue', '$accessvalue' )";
$resultaccess = mysqli_query($con, $sqlloc) or die (mysqli_error($con));

Answer 2

You should insert foreign key to your location table to maintain relation you have created before.
Your query for location should be like
`$sqlloc = "INSERT INTO 'location' ('ID', 'name', 'water', 'footer', 'access') VALUES ('".$_SESSION['loc_id']."', '$name', 'water-related-id', 'fodder-related-id, 'access-related-id)";`
Anyway, primary key for a table should be 'id' column and foreign key should be other table's primary key, so your constraints should be
ALTER TABLE 'location'
ADD CONSTRAINT 'location_ibfk2' FOREIGN KEY ('water') REFERENCES 'watercondition' ('id'),
ADD CONSTRAINT 'location_ibfk3' FOREIGN KEY ('fodder') REFERENCES 'foddercondition' ('id'),
ADD CONSTRAINT 'location_ibfk4' FOREIGN KEY ('access') REFERENCES 'accesscondition' ('id');
And, your foreign key data type should match referenced table's key which in your case is int(11).
Are you doing this for college assignment?

Answer 3

Change your last insert statement like below:
"INSERT INTO `location` (`ID`, `name`,`water`,`fodder`,`access`) VALUES
('".$_SESSION['loc_id']."', '$name','$watervalue','$foddervalue','$accessvalue')";
Explanation:
ERD:
Look the yellow marked tables are the parent tables and your location table is child table.
So according to MySQL you cannot add or update a child table row (location) by some foreign key values which don't exist in corresponding parent table(s).
In your first three insert statements you are updating your three parent tables which is fine.
But in your final insert statement you are trying to update your child table where you provide with no values for those foreign keys which is an exception.
SQL FIDDLE
Reference

Answer 4

Twothings to note here:
First of all, your insert is not working because you made all the foreign keys "NOT NULL" fields, so when you insert a record on the location table you must provide those values.
Second, you must make sure the fields that are foreign keys has the same data type on the original table. In your code you are using INT(11) id fields on the original tables but are using VARCHAR(25) on the location table and you are linking to the field holding the value instead of linking to the primary key (id) field of the referenced table.
Kind regards,
Daniel