如何为 PHP 自定义我自己的状态消息？ [等候接听]

Question

When I am at other websites and I create a new account, I get an alert style dialog box telling me if my info was successfully sent, while staying on the same webpage I created my account on. That dialog box has an "OK" button to select that closes the dialog box and resets the HTML form back to empty fields.

I want to do this on my website. Currently, I have an HTML form that captures email addresses. When I click SUBMIT, the PHP file takes the email address, verifies the email, uploads it into mysql, then opens a new browser window echoing the status of the PHP processes that were just run.

I do not want PHP opening a browser window where it displays the status of the processes. I simply want a dialog box to open in the same webpage that the form is on stating that the email address was successfully uploaded, or not successfully uploaded, or that the email address is invalid. Either way, no opening a second window, and an alert style dialog box that displays the status with an "OK" button to close the dialog box, and reset the form with blank fields.

I have spent 2 days searching the internet with various wording and I have explored the library here, but I cannot find help on this.

I am thinking my question has a standard solution because this is how most websites work when starting an account. I have probably been using incorrect terminology when searching. Here is my php code to upload the HTML form data into mysql:

<?php

    //MySQL/MariaDB connection block with the server location (localhost), the database name, and the database username and password.

    $email_a = $_POST['emailDesktop'];
    $email_b = $_POST['emailVerifyDesktop'];

    if (filter_var($email_a, FILTER_VALIDATE_EMAIL)) {
        echo "Email address '$email_a' is considered valid.\n";
}
    if (filter_var($email_b, FILTER_VALIDATE_EMAIL)) {
        echo "Email address '$email_b' is considered valid.\n";
}   else {
        echo "Email address '$email_b' is considered invalid.\n";
}


    $hostname = "xxxxxxxxxxxxxx";
    $username = "xxxxxxxxxxxxxx";
    $password = "xxxxxxxxxxxxxx";
    $db = "xxxxxxxxxxxxxx";

    //Then we add a section to connect to the database, and give an error if the connection fails:

    $dbconnect=mysqli_connect($hostname,$username,$password,$db);

    if ($dbconnect->connect_error) {
      die("Database connection failed:" .$dbconnect->connect_error);
    }


    //Next, we "translate" the data from the HTML form into PHP variables:

    if(isset($_POST['submit'])) {
        $emailDesktop=$_POST['emailDesktop'];




        //Then insert this information into the database:

        $query = "INSERT INTO members (emailDesktop)
        VALUES ('$emailDesktop')";


        //Add a nested if statement which will give an error message if the process fails, or thank the user for their review if it succeeds:

        if (!mysqli_query($dbconnect, $query)) {
        echo 'An error occurred when submitting.';
        } else {
        echo 'Upload successful.';


        }
    }

Answer 1

You can use ajax to submit your form data to the server and a modal to alert the end-user about the response.

The following code is an example of submitting data via ajax, using jQuery :

$("button").click(function(){
  $.ajax({
    url: "verify_email.php",
    data: your_data,
    success: function(result){
      showalert(result);
    }
  });
});

Refer to w3schools to learn how to make a CSS/JS modal.