php add songID in mysql - php

Question

I have a table and I want to add songs into my database. Every song has its own songID , so when I try to add a song it also has to update the songID. at this moment it didn't work.
if($action == "add-songs"){
$db = mysqli_connect($host, $user, $pass,$database);
if($_GET['action3'] == "2"){
mysqli_query($db, "INSERT INTO songs (id, songName, songID) VALUES (0, '".$_GET['songname']."', '".$_GET['id']."')");
header("Location: /?action=show-songs");
}
$h = "";
$h.= "";
$h.= "<form><input type='hidden' name='action' value='add-songs'><input type='hidden' name='action3' value='2'><input type='hidden' name='id' value='id'><table class='table table-striped'>";
$h.= " <tr>";
$h.= " <td><b>Nummer</b></td>";
$h.= " <td><input type='text' name='songname' class='form-control' placeholder='Naam'></td>";
$h.= " </tr>";
$h.= " <tr>";
$h.= " <td colspan='2'><input class='btn btn-primary' type='submit' value='UPDATE'></td>";
$h.= " </tr>";
$h.= "</table></form>";
echo $htop;
echo $h;
echo $hbot;

Answer 1

The problem is because of the following <input> element,
<input type='hidden' name='id' value='id'>
You're using the same string value id as id column value for all of your songs.
Now look at your songs table structure, id column is of type INT, which means you're inserting a string value in the column of type integer. And this is why your business logic is failing.
The solution is, don't make the above input element hidden. Explicitly put the song id corresponding to the particular song name. So change the above <input> element like this:
<input type='text' name='id' placeholder='id'>