How to pass a jQuery variable to PHP code?

Question

I have reframed my Question:

I have a drop down box which is populated with a list of users.

 <a href="#" class="Share<?=$r['Form']['id'];?>">Share</a>
    <form action="/FormBuilder/index.php/forms/share/<?=$r['Form']['id']?>/<?=$userid?>" method="post" id="shareform<?=$r['Form']['id'];?>" class="shareform">
            <p>Select the users with whom you want to share the Form</p>
            <select id="userList" name="userList" multiple>
                    <?php foreach($users as $user){  ?>
                            <option value=<?=$user['User']['id'];?>><?=$user['User']['name'];?></option>        
                    <?php }?>
            </select> 
            <input type="submit" class="button" value="Share"/>
    </form>

When I select the user and give submit, the Form's access field in the database is updated to "Shared".

Now I need to display the Form name that has been shared in the home page of the user whom I selected. This is the code for displaying the Form names that have been marked as "shared".

 <?php if (!empty($formsShared)){   
 foreach ($formsShared as $shared): 
  echo $shared['Form']['name']; 
 endforeach; 
 }
 else{
  echo "There are no shared Forms !"; 
 }?>

I need to get the selected values in the drop down box, so that I can have another for loop to display the Shared form only in the selected users home page. How do I get the option:value?

Answer 1

I don't think this is possible within the page. You need to submit in the form or pass it via an xhr post/get.

Then you could asses it in PHP via the regular parameters.

Answer 2

Any time you want to get data from JavaScript back into PHP, you are looking at a request+response.

XHR would be the way to go. You obviously aren't obligated to send back anything if all you're doing is receiving information or updating say...session state.

Answer 3

You will want to send them a PHP script using AJAX.

$.post("file.php", { data: selected_value[i] } );

and in your PHP script:

$data = $_POST['data'];

If the drop down is just one part of the form, then you create a hidden form field, with the value as *selected_value[i]*.

Answer 4

hello everybody

                       You will want to send them a PHP script using AJAX.
                       $.post("file.php", { data: selected_value[i] } );
                       and in your PHP script:
                       $data = $_POST['data'];

with this script, give me a simple example valid.

thank you...