Select 表格 - 在第三个表中链接两个表

Question

It's my first time here so please, be gentle...

My name is Maxime and I'm trying to learn PHP for few weeks now. I searched a lot on stack and Internet but I didn't find the solution, so I'm sorry if this is "basic" or if I don't explain very well. I'm looking for help!

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This week, I'm trying something: I have 2 tables:

BARLIST

|-----------+--------+----------+-----+---------|
|   barId   | barName| Adresse  | lat | long    |
|-----------+--------+----------+---------------|
|     1     |   Mike |50 street | 50  |  1.5    |
|     2     |   John |51 street | 30  |  45     |
|     3     |   Beth |52 street | 26  |  87     |
|-----------+--------+----------+---------------|

BEERLIST

|--------+----------+----------|
| beerId | beerName | typeBeer |
|--------+----------+----------|
|   1    |   Leffe  | blond    |
|   2    |   Affli  | brune    |
|   3    |   Affli  | blanche  |
|--------+----------+----------|

BARBEER - The linking table I would like

|--------+----------+----------|
| Id     | beerId   | barId    |
|--------+----------+----------|
|   1    |   1      | 1        |
|   2    |   2      | 1        |
|   3    |   3      | 2        |
|--------+----------+----------|

The idea is to let the user choose a bar and a beer from the tables through a select form, to add them into the third table barbeer . So I've trying few things and recently here what I have.

Form with select content

<form method="post" action="">
            <label class="first-column ">Choose a bar</label><div class="second-column">
                <select name="test">
                    <option>Select</option>
                    <?php

                    $sqli = "SELECT * FROM barList";
                    $result = mysqli_query($con, $sqli);
                    while ($row = mysqli_fetch_assoc(@$result)) {
                        ?>
                        <option value="<?php echo $row['barId'];?>"><?php echo $row['barName'];?></option>';
                        <?php
                    }


                    ?>
                </select>

                <label class="first-column ">Choose a beer</label><div class="second-column">
                    <select name="test1">
                        <option>Select</option>
                        <?php

                        $sqli = "SELECT * FROM beerList";
                        $result = mysqli_query($con, $sqli);
                        while ($row = mysqli_fetch_assoc(@$result)) {
                            # code...?>
                        <option value="<?php echo $row['beerId'];?>"><?php echo $row['beerName'];?></option>';
                      <?php
                        }


                        ?>
                    </select>
            <input type="submit" name="submit_4" value="Submit" />

It works great and I can see on my form the name of the beer and the bar. But now I would like to add the selection of the user into a third table, so I tried this:

if(isset($_POST['submit_4'])){
    if(empty($errors)){
        try
        {
            $pdo = new PDO('mysql:host=localhost;dbname=whereismybeer;charset=utf8', 'root', '');
        }
        catch(Exception $e)
        {
            die('Erreur : '.$e->getMessage());
        }
        echo $_POST['test'];
        echo $_POST['test1'];
        
        $req= $pdo-> prepare("INSERT INTO barbeer JOIN barList, ON barbeer.barId = ? JOIN beerList, ON barbeer.beerId = ?");
        $req->execute([$_POST['test'], $_POST['test']]);
        
        die ('Yeah, It worked!');

    }}

Also I tried with a variable like this

$var = $_POST['test];
INSERT INTO [...] barbeer.barID = $var;

I had no error messages, just the "Yeah, It worked." but nothing on the barbeer table... So I'm stuck, Could you help me to find a solution? ideas or anything useful?

Thank you very much for the answer and the help!

EDIT: In another page, in the future, the idea will be to ask "where can I drink this beer?", and the website to show the list of the bar where I could drink this beer. That's why I thought a Joint would be necessary.

EDIT 2: Ok, so my understanding of the JOIN function was wrong. I don't need to call it now. I just need to put the beerID and the barID into the third table with a classic INSERT.

INSERT INTO barbiere (barName, biereName) VALUES (?, ?);

Answer 1

You don't need to use join. You have the values of beerId, barId from the form. So just do a normal insert.

The statement should be.

INSERT INTO barbeer (beerId, barId) VALUES (?, ?);