Trying to join certain values in different tables in mysql - php

Question

i'm very new to mysql and I am trying to create a database that can store users emails and passwords on one table and the values they input on another table, how do I join the tables to make sure that the inputted values are linked to the correct user. This is the code I've been using but it won't allow the value to be stored while the foreign key is run, but if I remove the foreign key I can store the value. Please help.
CREATE TABLE IF NOT EXISTS `data` (
`user_id` int(11) NOT NULL AUTO_INCREMENT,
`email` varchar(51) NOT NULL,
`password` varchar(15) NOT NULL,
PRIMARY KEY (`user_id`),
UNIQUE KEY `email_UNIQUE` (`email`)
)
CREATE TABLE IF NOT EXISTS `gluco` (
`G_id` int(11) NOT NULL AUTO_INCREMENT,
`bloods` decimal(4,2) NOT NULL,
`time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`user_id` int(11) NOT NULL,
FOREIGN KEY (`user_id`) REFERENCES `data`(`use_id`),
UNIQUE KEY `G_id_UNIQUE` (`G_id`)
)
<?php
include('db.php');
if (!isset($_POST['reading'])) { //checking if user has entered this page directly
include('contactus.php');
} else {
if (isset($_POST['reading'])&&$_POST['reading']==""||!isset($_POST['reading'])) {
$error[] = "fill in your blood/glucose";
}
$reading = mysql_real_escape_string($_POST['reading']);
$sql = "SELECT * FROM gluco WHERE bloods = '$reading'";
if(isset($error)){
if(is_array($error)){
echo "<div class=\"error\"><span>please check the errors and refill the form<span><br/>";
foreach ($error as $ers) {
echo "<span>".$ers."</span><br/>";
}
echo "</div>";
include('contactus.php');
}
}
if(!isset($error)){
$sreading=mysql_real_escape_string($_POST['reading']);
$sip=mysql_real_escape_string($_SERVER['HTTP_HOST']);
$save = mysql_query("INSERT INTO `gluco` ( `bloods` )VALUES ('$sreading')");
if($save){
echo "<div class=\"success\"><span>Your reading has been successfully stored</span><br/></div>";
} else {
echo "<div class=\"warning\"><span>Some Error occured during processing your data</div>";
}
}
}
?>

Answer 1

your code is correct in its logic. But theres an error on the referenced column name:
CREATE TABLE IF NOT EXISTS `data` (
`user_id` int(11) NOT NULL AUTO_INCREMENT,
`email` varchar(51) NOT NULL,
`password` varchar(15) NOT NULL,
PRIMARY KEY (`user_id`),
UNIQUE KEY `email_UNIQUE` (`email`)
)
CREATE TABLE IF NOT EXISTS `gluco` (
`G_id` int(11) NOT NULL AUTO_INCREMENT,
`bloods` decimal(4,2) NOT NULL,
`time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`user_id` int(11) NOT NULL,
FOREIGN KEY (`user_id`) REFERENCES `data`(`user_id`),
UNIQUE KEY `G_id_UNIQUE` (`G_id`)
)
and on this line:
$save = mysql_query("INSERT INTO `gluco` ( `bloods` )VALUES ('$sreading')");
you are not setting the user_id in your insert statement, so, the foreign key will not work and the insert will not be made. So, you'll need to have the user id stored in a variable (since i don't know the context and the scope in the code, i can't help you setting this variable). So, your code should be like that:
$save = mysql_query("INSERT INTO gluco (bloods, user_id)VALUES ('$sreading', $user_id)");