I want to insert the foreign key of my customer_id in my feedback table. How should I do it?
customer table:
customer_id
coach_id
customer_name
feedback table:
feedback_id
feedback1
feedback2
customer_id
I want to do it where after user login, user insert information of the feedback it will automatically register the customer id.
This is my code after I login and want to register feedback:
<?php
session_name ('YourVisitID');
session_start();
$page_title = 'Feedback';
include('./header4.html');
//remember to delete.
echo "{$_SESSION['customer_name']}";
?>
<section id="main" class="wrapper">
<div class="container">
<form action = "feedback.php" method="post">
<div class="row uniform 50%">
<div class="6u 12u$(xsmall)">
<input type="text" name="weight" placeholder="Weight"
required autofocus/>
</div>
<div class="6u$ 12u$(xsmall)">
<input type="text" name="height" placeholder="Height"
required autofocus/>
</div>
<div class="6u 12u$(xsmall)">
<input type="text" name="water" placeholder="Water Level%"
required autofocus/>
</div>
<div class="6u$ 12u$(xsmall)">
<input type="text" name="body_fat" placeholder="Body Fat%"
required autofocus/>
</div>
<div class="6u 12u$(xsmall)">
<input type="text" name="calorie" placeholder="Calorie"
required autofocus/>
</div>
<div class="6u$ 12u$(xsmall)">
<input type="text" name="visceral" placeholder="Visceral Fat Level%"
required autofocus/>
</div>
<p><input type="submit" name="submit" value="Register" /></p>
<input type="hidden" name="submitted" value="TRUE" />
</div>
</form>
</div>
</section>
<?php
if(isset($_POST['submitted'])) {
require_once ('mysql_connect3.php');
function escape_data ($data){
global $dbc;
if (ini_get('magic_quotes_gpc')){
$data = stripslashes($data);
}
return mysql_real_escape_string(trim($data), $dbc);
}
$error = array();
$weight = escape_data($_POST['weight']);
$height = escape_data($_POST['height']);
$water = escape_data($_POST['water']);
$calorie = escape_data($_POST['calorie']);
$visceral = escape_data($_POST['visceral']);
$fat = escape_data($_POST['body_fat']);
mysqli_close($con);
header("location: add_user.php?remarks=success");
if (empty ($errors)) {
$query ="SELECT * FROM feedback WHERE weight ='$weight'";
$result = mysql_query($query);
if (mysql_num_rows($result) == 0) {
$query = "INSERT INTO feedback (weight, height, body_fat, water, calorie, visceral, feedback_date) VALUES
('$weight', '$height', '$water', '$calorie', '$visceral','$fat', NOW() )";
$result = @mysql_query ($query);
if ($result) {
echo '<script>
alert("Your feedback has been save");
</script>';
include ('./footer.html');
exit();
}else{
echo '<script>
alert("<h1 id="mainhead">System Error</h1>
<p class="error">You could not give feedback due to a system error.
We apologize for any inconvenience.</p>");
</script>';
echo '<p>'. mysql_error() . '<br /><br />Query: ' . $query . '</p>';
include ('./footer.html');
exit();
}
}
}else{
echo '<script>
alert("<h1 id="mainhead">Error!</h1>
<p class="error">Please try again.</p>");
</script>';
}
mysql_close();
}
?>
<?php
include ('./footer.html');
?>
I think you should define in sql table schema.
After you define foreign key customer_id in feedback table
You can use PHP to select data from that table
When a customer logs in, just store its customer_id in the session, just as you store its customer_name at the moment. When the customer submits the feedback form, you simply provide the customer_id from the session.
$query = "INSERT INTO feedback (weight, height, body_fat, water, calorie, visceral, feedback_date, customer_id) VALUES
('$weight', '$height', '$water', '$calorie', '$visceral','$fat', NOW(), ".$_SESSION ['customer_id']. ")";
$result = @mysql_query ($query);
Couple of notes: