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> Dropdown Menu with mySQL backend
zer0uk
post Mar 5 2020, 11:09 AM
Post #1





Group: Members
Posts: 2
Joined: 5-March 20
Member No.: 27,219



Hello,

must admit i'm very new to this and just a hobiest ,
I am creating a page which will be used for logging bike ride information.

With a mysql database i have two tables "tbl_rides" & "tbl_courses" tbl_rides is the main table for storing information and has a foreign key associating column "course_id" with course_id in the tbl_courses

What i want to do on a HTML/PHP page is display a drop down box that has the contents of "course_name" from the tbl_courses table but when the user selects it and submits the record is written to the main table "tbl_rides" using FK "course_id" so display the friendly name the foreign key but the record is written to "tbl_rides" use the associated field course_id

my work in progress code is below

<?php

// database connect

include('../db_connect.php');



//-------- Check Data -----------

//Check if Data is sent & validate

if(isset($_POST['name_submit'])){

if(empty($_POST['name_submit'])){

Echo 'No Course passed';

} else {

$var_course_id = mysqli_real_escape_string($conn, $_POST['course_id']);



//Create sql

$sql = "INSERT INTO tbl_rides(course_id) VALUES ('$var_course_id')";

//Save to DB and check

if(mysqli_query($conn, $sql)){

//success




} else {

echo 'query error: ' . mysqli_error($conn);

}

}



}



//end of checking



?>




<!DOCTYPE html>

<html>





<form class="white" action="dropdown.php" method="POST">

<select name="course_id">

<?php



// query to create course_id dropdown

$resultset = $conn->query("SELECT course_id FROM tbl_rides");

while($rows = $resultset->fetch_assoc())

{

$var_course_id = $rows['course_id'];

echo "<option value='$var_course_id'>$var_course_id</option>";

}

?>



</select>

<div class="centre" >

<input type="submit" name="name_submit" value="Submit my information !" />

</div>

</form>




</html>
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