Foreign Key Failure in MySQL - php

Question

I have created a database composed of three tables. This is my query in creating my tables with Foreign Key.
CREATE TABLE reporter
(
reporterid INT NOT NULL AUTO_INCREMENT,
firstname VARCHAR(1000) NOT NULL,
lastname VARCHAR(100) NOT NULL,
PRIMARY KEY (reporterid)
);
CREATE TABLE flood
(
floodid INT NOT NULL AUTO_INCREMENT,
address VARCHAR(500) NOT NULL,
description VARCHAR(1000) NOT NULL,
dateofflood DATE NOT NULL,
timeofflood INT NOT NULL,
PRIMARY KEY (floodid)
);
CREATE TABLE reports
(
reportid INT NOT NULL AUTO_INCREMENT,
timereport NODATATYPE NOT NULL,
datereport DATE NOT NULL,
rid INT NOT NULL,
fid INT NOT NULL,
PRIMARY KEY (reportid),
FOREIGN KEY (rid) REFERENCES reporter(reporterid),
FOREIGN KEY (fid) REFERENCES flood(floodid)
);
I created a system in order for me to add records/row on my database through PHP. This is my code:
<?php
mysql_connect("localhost", "root", "") or die("Connection Failed");
mysql_select_db("flooddatabase")or die("Connection Failed");
$description = $_POST['description'];
$address = $_POST['address']; // Make sure to clean the
$dateofflood=$_POST['dateofflood'];
$timeofflood=$_POST['timeofflood'];
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$dateofreport=$_POST['dateofreport'];
$timeofreport=$_POST['timeofreport'];
$query = "INSERT into flood(address,description,dateofflood,timeofflood) values ('$address','$description','$dateofflood','$timeofflood')";
$query2 = "INSERT into reporter(firstname,lastname) values ('$firstname','$lastname')";
$query3 = "INSERT into reports(dateofreport,timeofreport) values ('$dateofreport','$timeofreport')";
if(mysql_query($query))
if(mysql_query($query2))
if(mysql_query($query3))
{
echo "";
} else
{
echo "fail";
}
?>
The result that I am getting is fine. It's just that, in my REPORTS table, there is no foreign key that is being generated. For example I input something on my reporter table and flood table, the foreign key 'rid' and 'fid' has no values that references to both tables. Need help thank you.

Answer 1

Get the just inserted Primary key value from flood table insert
query. And store it to a variable say $f_id;
Get the just inserted primary key value from reporter table insert
query and store it to a variable say $r_id;
Now Make your last insert statement like below:
"INSERT into reports(dateofreport,timeofreport,rid,fid) values ('$dateofreport','$timeofreport',$r_id,$f_id)";
I am not giving you a direct copy paste solution.
If you need to know how to get the last inserted id by executing an insert query then look at this link

Answer 2

there is no foreign key that is being generated
I'm not entirely sure what you even mean by that. Foreign keys aren't "generated". Primary keys can be, which you do:
reporterid INT NOT NULL AUTO_INCREMENT
(as well as for your other two tables)
the foreign key 'rid' and 'fid' has no values
Well, look at your query:
INSERT into reports(dateofreport,timeofreport) values ...
Where do you insert values for rid and fid? I'm actually pretty surprised this query works at all, since those columns don't allow NULL values:
rid INT NOT NULL,
fid INT NOT NULL,
(Though your column names also don't line up, so I find it likely that the code you're showing isn't actually the code you're using...) That point aside however, the fact still remains that if you want a value in those fields then you have to put a value in those fields:
INSERT into reports(dateofreport,timeofreport,rid,fid) values ...
After each query, you can get the last generated identifier from mysql_insert_id():
$last_id = mysql_insert_id();
Use that to then populate the values being inserted as foreign keys in subsequent queries.
Also worth noting, the mysql_* libraries are long since deprecated and have been replaced with mysqli_ and other libraries such as PDO. I highly recommend you upgrade to a current technology, since what you're using isn't supported by any vendor.
Additionally, and this is very important, your code is wide open to SQL injection attacks. This basically means that you execute any code your users send you. You should treat user input as values, not as executable code. This is a good place to start reading on the subject, as is this.