How to use same php function for different button clicks

Question

Here is a simple PHP form with a button..

<form method="POST">
  <div class="mb-3">
       <button  type='button'  id ="btnnew1" class="btn btn-info" >submit</button>
       <p></p>
  </div>
</form>

Here is the Jquery functions which executes a PHP file.

$(document).ready(function(){
    $("#btnnew1").click(function(e){
        if(!confirm('Are you sure?')){
            e.preventDefault();
            return false;
        }
        else{
          $.ajax({
            url: 'test.php',
            success: function(data) {
                $("p").text(data);
             }
          });
          
        }
    });
});

And the test.php is as follows,

<?php

echo 'Button1 clicked' 
?>

My question is how to modify my test.php if I have multiple buttons.

As an example,

<form method="POST">
      <div class="mb-3">
           <button  type='button'  id ="btnnew1" class="btn btn-info" >submit</button>
           <p></p>
      </div>
<div class="mb-3">
           <button  type='button'  id ="btnnew2" class="btn btn-info" >submit</button>
           <p></p>
      </div>
<div class="mb-3">
           <button  type='button'  id ="btnnew3" class="btn btn-info" >submit</button>
           <p></p>
      </div>
    </form>

Result should be,

If btnnew1 clicks--->echo("Button1 clicked);
If btnnew2 clicks--->echo("Button2 clicked);
If btnnew3 clicks--->echo("Button3 clicked);

Update:

What If I need to run three different php functions(no any pattern)?

Ex:
    If btnnew1 clicks--->
     sleep(5)
     echo("Button1 clicked);

If btnnew2 clicks--->
     sleep(15)
     echo("Button2 clicked by user);
If btnnew3 clicks--->
     sleep(35)
     echo("Button3 clicked by user);

Answer 1

In here I am changing little bit your default settings. This will help you as I can understand. You can try as below,

1)Change your button into input type..I have added some inline CSS as well. If you don't like you may neglect it...

<input type="button" style="background-color: #3CBC8D;padding:3px;"   class="button" name="fcn1" value="Update My Status"/>


<input type="button" class="button" style="background-color: #3CBC8D;padding:3px;" name="fcn2" value="Update My Status" />

2. Then go to jquery and use as below, success function change as you wish. Here I have used an alert box...

      $(document).ready(function(){
          $('.button').click(function(){
   
            if(!confirm('Are you sure?')){
                e.preventDefault();
                return false;
            }
            else{
            var clickBtnValue = $(this).attr('name');
            var fetchdata= 'testme.php',
            data =  {'action': clickBtnValue};
            $.post(fetchdata, data, function (response) {
                // Response div goes here.
   
                alert("Updated successfully -"+response);
   
            });
            }
   
        });
   
    });
   

3. Finally change your testme.php as follows,

    if (isset($_POST['action'])) {
        switch ($_POST['action']) {
            case 'fcn1':
                fcn1();
                break;
            case 'fcn2':
                fcn2();
                break;
        }
    }
   
    function fcn1() {
   
        echo 'Button1 clicked';
        exit;
    }
   
    function fcn2() {
        sleep(5);
        echo 'Button2 clicked';
        exit;
    }
   

Answer 2

Set the name to each button: Button 1

Send data using ajax: Get button text using e.target.text and send using POST method.

$.ajax({
  type: 'POST',
  url: 'test.php',
  data: { buttonTitle: e.target.name},
  success: function(data) {
      $("p").text(data);
    }
});

php: Inside php use $_GET to get the data which we send from the frontend.

if(isset($_POST['buttonTitle'])) {    
  $buttonTitle = $_POST['buttonTitle'];
  echo $buttonTitle . " clicked";
}