How can I run a php function in a jquery function?

Question

I want to run a PHP function once I click a confirm button in a jquery modal box. Here is the code:

<?php
if(isset($_POST['submitBackground'])){
    $employee = new Employee();
    $employee->build($_POST);

    $confirm = true;
}

?>

    <?php if($confirm){ ?>
<script type="text/javascript">


$(document).ready(function(){

    // Demo modal
        function openModal()
        {
            $.modal({
                content: '<p>Please make sure that all the information is correct</p>'+

                          '<ul class="simple-list with-icon">'+
                          '    <li>First Name:</li>'+
                          '    <li>Last Name:</li>'+
                          '    <li>Address:</li>'+
                                                  '    <li>City:</li>'+
                                                  '    <li>State:</li>'+
                                                  '    <li>Zip Code:</li>'+
                                                  '    <li>Position:</li>'+
                                                  '    <li>Social Security #:</li>'+
                                                  '    <li>Drivers License:</li>'+
                                                  '    <li>Drivers License State:</li>'+
                          '</ul>',
                title: 'Confirm Application',
                maxWidth: 500,
                buttons: {
                    'Confirm': function(win) { openModal(); },
                    'Cancel': function(win) { win.closeModal(); }
                }
            });
        }

    // Demo modal
        openModal();

});
</script>

<?php } ?>

Then, if the person presses "Confirm" I want to run the "$employee->save()" function. How can I get that done?! Thank you!

If you want my advise, you should learn the difference between server-side languages and client side ones, BEFORE even hearing about jQuery ;). — Christian
– Christian, Commented Jan 25, 2011 at 10:10

Amit Battan · Accepted Answer · 2011-01-25 10:09:42Z

1

directly you cannot run you can call using ajax

EDIT : Call it in ajax page and at the same time you block the page using process bar

edited Jan 25, 2011 at 10:09

answered Jan 25, 2011 at 10:04

Amit Battan

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Comments

Yoram de Langen · Accepted Answer · 2011-01-25 10:12:06Z

0

You should send a ajax request to a file where you use your PHP class $employee->save()

<script type="text/javascript">

function ButtonSaveClick()
{
 $.ajax({
   url: '[.. to your php file ..]',
   data: $('#employeeFrm').serialize(), //using serialize to POST every field (<input />)
   dataType: 'json', //only if you wanna return with JSON
   success: function(data) {
  // callback when done posting
  //when doing json
  if(data.status)
  {
   alert(data.message);
   // refreshList() or somthing
  }
  else
  {
   alert(data.message);
  };
   }
 });
}

$(document).ready(function(){

    // Demo modal
 function openModal()
 {
  $.modal({
   content: '<p>Please make sure that all the information is correct</p>'+
   '<form id="employeeFrm">'+  
   '<ul class="simple-list with-icon">'+
   '    <li>First Name:</li>'+
   '    <li>Last Name:</li>'+
   '    <li>Address:</li>'+
   '    <li>City:</li>'+
   '    <li>State:</li>'+
   '    <li>Zip Code:</li>'+
   '    <li>Position:</li>'+
   '    <li>Social Security #:</li>'+
   '    <li>Drivers License:</li>'+
   '    <li>Drivers License State:</li>'+
   '</ul>'+
   '</form>',
   title: 'Confirm Application',
   maxWidth: 500,
   buttons: {
    'Confirm': function(win) { ButtonSaveClick(); },
    'Cancel': function(win) { win.closeModal(); }
   }
  });
 }

    // Demo modal
    openModal();

});
</script>

So should my php file i use in my ajax post:

include_once "[dbfile]";
include_once "[employeeClass is stored]";

$employee = new Employee();
if($employee->build($_POST))
{
 //when successfully build
 //return json way
 echo json_encode(array('status'=>true, 'message'=>'Successfull')); exit;
 //return non json way
 echo 'Successfull'; exit;
}
else
{
 //when failed with building
 //return json way
 echo json_encode(array('status'=>false, 'message'=>'failed')); exit;
 //return non json way
 echo 'failed'; exit;
};

answered Jan 25, 2011 at 10:12

Yoram de Langen

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5 Comments

raygo Over a year ago

Hi. Thank you for your response!! I did what you said and I receive a "Successful" message. But when I put the $employee->save() after the "when successfully build", I dont get any message and the records dont get saved either. Do you have any idea why is that? Thanks!

Yoram de Langen Over a year ago

Did you put it before or after ` echo json_encode(array('status'=>true, 'message'=>'Successfull')); exit;` or echo 'Successfull'; exit; ?? because when you do Exit; to stop everything whats after exit; php.net/manual/en/function.exit.php

raygo Over a year ago

I tried it both ways and it doesnt work. I actually fixed some things and now im getting a "failed" and not a successful. I tried for the Modal to open when I click on the submit button, but I also have the validate action I guess in that button, so when I do an onclick="openModal()", neither the validate nor the modal work.

raygo Over a year ago

And also, if I do it like that, then how will I be able to put the values in the modal "first name: John, last name: Smith" etc... in order to confirm it? Sorry for the many questions!! Thanks for everything!

raygo Over a year ago

I did it! y just had to add type:'Post' ! Thank you!

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