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Datasbase value check and action if it does, PHP

Sam Vorst asked:

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Hello all,

I have an question, I have PHP code to create a unique ID but I want to be sure it is unique. If not it has to do an action: create a new ID that is not exists already.
This is my code:
<?php

    function random_strings($length_of_string)
    {

        $str_result = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz';

        return substr(str_shuffle($str_result), 0, $length_of_string);
    }

    ob_start();

    echo random_strings(10);

    $Output = ob_get_contents();

    ob_end_clean();



    $conn = mysqli_connect("localhost", "probleem_uniqid", "Password", "DBname");

    if($conn === false){
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }

    $Name = $_POST['Name'];
    $Email = $_POST['Email'];
    $Phone = $_POST['Phone'];
    $Problem = $_POST['Problem'];

    if ($stmt = mysqli_prepare($conn, "INSERT INTO ID (Id, Name, Email, Phone, Problem) VALUES (?, ?, ?, ?, ?)")) {
        mysqli_stmt_bind_param($stmt, "sssss", $Output, $Name, $Email, $Phone, $Problem);
        if (mysqli_stmt_execute($stmt)) {
          echo ' ';
          echo "<div id=\"PHP-output\"> Succesvol Verstuurd </div>";
        } else{
            echo ' ';
            echo "<div id=\"PHP-output\"> Sorry, Het kon niet worden verstuurd, probeer het later nog eens of neem contact op met de klantenservice. $sql.  </div>" . mysqli_error($conn);
        }
    }

    mysqli_close($conn);

?>


What I have tried:

I don't know very much about PHP, but I tried to follow yt tutorials or web tutorials but I can't come out to it.
Tags: PHP

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