CODEWITHSUNDEEP
phpbootstrap-modal
aadil
aadil

Reputation: 3

How to execute code after a button is clicked in modal

After the button "confirm" is clicked, I want to perform those operations. but my code doesn't run after the line `if (isset($_POST['confirm'])){..} i would really appreaciate your help. 

echo "<div class='container'>
    <form action='confirm.php' method='post'>
    <center><button type='button' name='confirm' value='confirm' class='btn btn-info btn-lg' data-toggle='modal' data-target='#myModal'>Confirm</button></center>
    </form>
    <div class='modal fade' id='myModal' role='dialog'>
    <div class='modal-dialog'>";


 if (isset($_POST['confirm'])){

        $query2 = "SELECT status FROM reserver WHERE picker='$picking' AND droper='$droping'";
        $result1 = mysqli_query($conn, $query2);
        $row1=mysqli_fetch_assoc($result1);
        $check_status=$row1['status'];
        if($check_status=='available')
        {

         $query3 = "UPDATE reserver SET status='reserved' WHERE picker='$picking' AND droper='$droping'";
         if(mysqli_query($conn, $query3)){



 echo " <div class='modal-content'>
    <div class='modal-header'>
      <button type='button' class='close' data-dismiss='modal'>&times;</button>
      <h4 class='modal-title'>Confirmation</h4>
    </div>
    <div class='modal-body'>
      <p>Confirmed Sucessfully!</p>
    </div>
    <div class='modal-footer'>
      <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
    </div>
  </div>";
         }
        }
      else  {
            echo " <div class='modal-content'>
    <div class='modal-header'>
      <button type='button' class='close' data-dismiss='modal'>&times;</button>
      <h4 class='modal-title'>Notice</h4>
    </div>
    <div class='modal-body'>
      <p>Unavailable Vehicle!</p>
    </div>
    <div class='modal-footer'>
      <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
    </div>
  </div>";
        }

 }

EDIT : now i have an else case for diaplaying unavailable vehicle

Upvotes: 0

Views: 4251

Answers (2)

Rahul Dev
Rahul Dev

Reputation: 175

Your solution will not open the modal with success message. You can try something like this Pastebin Link

<?php
$status = false;
if (isset($_POST['confirm'])) {

    $check_status = 'available';
    $query2 = "SELECT status FROM reserver WHERE picker='$picking' AND droper='$droping'";
    $result1 = mysqli_query($conn, $query2);
    $row1=mysqli_fetch_assoc($result1);
    $check_status=$row1['status'];
    if($check_status=='available')
    {
        $query3 = "UPDATE reserver SET status='reserved' WHERE picker='$picking' AND droper='$droping'";
        if(mysqli_query($conn, $query3)) {
            $status = true;
        }
    }
}
?>
<div class='container'>
    <form action='confirm.php' method='post'>
        <center><button type='submit' name='confirm' value='confirm'>Confirm</button></center>
        <button type='button' name='confirm' value='confirm' class='btn btn-info btn-lg' data-toggle='modal' data-target='#myModal' style="display:none;">Confirm</button>
    </form>
    <div class='modal fade' id='myModal' role='dialog'>
        <div class='modal-dialog'>
            <div class='modal-content'>
                <div class='modal-header'>
                    <button type='button' class='close' data-dismiss='modal'>&times;</button>
                    <h4 class='modal-title'>Confirmation</h4>
                </div>
                <div class='modal-body'>
                    <p>Confirmed Sucessfully!</p>
                </div>
                <div class='modal-footer'>
                    <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
                </div>
            </div>
        </div>
    </div>
</div>
<?php if( $status == true ): ?>
    <script>
        $('.btn.btn-info').trigger('click');
    </script>
<?php endif; ?>

Upvotes: 1

Mourad Karoudi
Mourad Karoudi

Reputation: 388

However what you want to do if you want to submit data and process it by php you need to use <input type='submit'/> not button.

<input type="button" /> work with javascript and will not submit a form. <input type="submit" /> will submit the form.

but AJAX is here to do this for you just you to learn some basics to do that and is much easier with jQuery

So here is example how that works

First thing you need is to create a simple HTML form

if you want append some data

<form method="POST" id="my_form">
<input type="hidden" name='test'/>
<input type="button" id="submit"/>
</form>

OR just create a simple button

<button id="click me" id="click_me"></button>

and then create a another HTML element to show user response status

<div id="response"></div>

Good. So now we finish our html code let's go to write some jQuery code

$(document).ready(function() {
$("#click_me").click(function() {
$.ajax({
url : "your_php_file.php", // file path you want to send data
type : "POST",
data : {},
success : function(data) {
     $("#response").html("Confirmed Sucessfully!");
}
});
});
});

And the last thing is your php code

<?php
if() {
// your php code
}else{
header("HTTP/2.0 400 Bad request");
}
?>

This is just a simple explain quickly how to handle AJAX request. You need to learn more about this to build secure and useful web application

Upvotes: 1

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