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i have to two tables

users table:
CREATE TABLE `users` (
  `id` int(11) NOT NULL,
  `fname` varchar(255) NOT NULL,
  `lname` varchar(255) NOT NULL,
  `username` varchar(255) NOT NULL,
  `email` varchar(255) NOT NULL,
  `password` varchar(255) NOT NULL,
  `phone` int(11) NOT NULL,
  `role` enum('admin','user') NOT NULL DEFAULT 'user',
  PRIMARY KEY (`id`),
  UNIQUE KEY `email` (`email`)

|id |fname | lname| username | email | password | phone |
----- ------ ------ ---------- ------------------ ---------- ----------
| 123| john | bale | john22 | | hashed | 0123456 |
| 20 | mike |Taylor| mike123 | | hashed | 456789 |

cars table:

  `vid` int(11) unsigned NOT NULL,
  `user_id` int(11) DEFAULT NULL,
  `model` varchar(255) NOT NULL,
  `reg_number` int(20) unsigned NOT NULL,
  `year` int(10) unsigned NOT NULL,
  `color` varchar(128) NOT NULL,
  PRIMARY KEY (`vid`),
  UNIQUE KEY `reg_number` (`reg_number`),
  KEY `user_car_id` (`user_id`),
  CONSTRAINT `user_car_id` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`)

| vid | user_id | model | reg_number  | year | color |
------ --------- -------- ------------- ------ -------
| 555 | NULL | toyota | 258 | 2000 | white |
| 099 | NULL | mazda | 852 | 2015 | gray |

what i want is, when the user 'for example' john his id=123 submit the car form his id appear in the cars table in the column user_id.

here is the car form in html:

<form action="" method="post" autocomplete="off" >
       <!-- Second row -->
       <div class="row pt-4">
           <div class="col-3">
         <label for="vehicle" class="form-label">Vehicle-ID:</label>
         <input type="number" class="form-control" name="vid" id="vehicle" >
       <div class="col-3">
         <label for="model" class="form-label">Vehicle Model:</label>
         <input type="text" class="form-control" name="model" id="model" placeholder="E.g. Toyota" >
       <div class="col-3">
           <label for="registration" class="form-label">Registration Number:</label>
           <input type="number" class="form-control" name="reg_number" id="registration" >

         <!-- Third row -->
         <div class="row">
           <div class="col-3">
         <label for="year" class="form-label">Year of Production:</label>
         <input type="number" class="form-control" name="year" id="year"  >
       <div class="col-3">
         <label for="color" class="form-label">Vehicle Color:</label>
         <input type="text" class="form-control" name="color" id="" >

           <div class="row">
               <div class="col-12 text-center">
             <button type="submit" class="btn btn-light my-3">


and here's the php insertion data code:


 $vid = mysqli_real_escape_string($mysql, $_POST['vid']);
 $model = mysqli_real_escape_string($mysql, $_POST['model']);
 $reg_number = mysqli_real_escape_string($mysql, $_POST['reg_number']);
 $year = mysqli_real_escape_string($mysql, $_POST['year']);
 $color = mysqli_real_escape_string($mysql, $_POST['color']);

 if (empty($vid)) {
  array_push($errors, "vehicle id is required!");
if (empty($model)) {
  array_push($errors, "vheicle model is required!");
if (empty($reg_number)) {
  array_push($errors, "registeration number is required!");
if (empty($year)) {
  array_push($errors, "year of production is required!");
if (empty($color)) {
  array_push($errors, "vheicle color is required!");

  if (!count($errors)) {
    $carExists = $mysql->query("select vid, reg_number from cars where vid ='$vid' limit 1 ");

    if ($carExists->num_rows) {
      array_push($errors, "vheicle id already registed!");

  if (!count($errors)) {
    $carExists = $mysql->query("select vid, reg_number from cars where reg_number ='$reg_number' limit 1 ");

    if ($carExists->num_rows) {
      array_push($errors, "registeration number already registed!");

    $carq = "insert into cars (vid, model, reg_number, year, color) values ('$vid','$model','$reg_number','$year', '$color')";

    $_SESSION['logged_in'] = true;
    $_SESSION['car_id'] = $mysql->insert_id;
    $_SESSION['success_message'] = "car submitted successfully";
    header('location: carformm.php');

What I have tried:

i tried this query but didn't work:

$carq = "insert into cars (vid, model, reg_number, year, color, user_id) values ('$vid','$model','$reg_number','$year', '$color','$_POST[user_id]')";

this is the error:

Warning**: Undefined array key "user_id" in **C:\\xampp\\htdocs\\flex-tut\\Carformm.php** on line **53**
> **Fatal error**: Uncaught mysqli_sql_exception: Cannot add or update a child row: a foreign key constraint fails (\`carcare\`.\`cars\`, CONSTRAINT \`user_car_id\` FOREIGN KEY (\`user_id\`) REFERENCES \`users\` (\`id\`)) in C:\\xampp\\htdocs\\flex-tut\\Carformm.php:54 Stack trace: #0 C:\\xampp\\htdocs\\flex-tut\\Carformm.php(54): mysqli-\>query('insert into car...') #1 {main} thrown in **C:\\xampp\\htdocs\\flex-tut\\Carformm.php** on line **54**

what can i do to make it work?
Updated 6-Feb-23 0:54am

1 solution

Your error is created because you have never posted the user_id to your form but a reference were made to it.

The easiets way for this is to store your user info in a session which can be called upon anytime during their active session on any page it is required.

Read more about sessions, creating a session, closing a session etc from THIS PHP manual.

Some basic sample code usage will look like this -
// page1.php


echo 'Welcome to page #1';

$_SESSION['favcolor'] = 'green';
$_SESSION['animal']   = 'cat';
$_SESSION['time']     = time();

// Works if session cookie was accepted
echo '<br /><a href="page2.php">page 2</a>';

// Or maybe pass along the session id, if needed
echo '<br /><a href="page2.php?' . SID . '">page 2</a>';
//After viewing page1.php, the second page page2.php will magically contain the session data.
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