Question

Let’s say that I have a file (file1.php) with a simple form with the action attribute:

echo'
<form action="foo.php" method="post">
Name:  <input type="text" name="username" /><br />
Email: <input type="text" name="email" /><br />
<input type="submit" name="submit" value="Submit me!" />
</form>';
$another_var = $user_id+5; 

Let’s say the foo.php looks something like this:

$sql ... SELECT user_id, username... WHERE username = $_POST['username']; //or so
echo 'We got the user ID. it is in a variable!';
$user_id = $row['user_id'];

As you see, I need the variable $user_id made in foo.php actually to be used in the main file file1.php.
Is there any way to do this? I though that return $user_id would work but I was wrong :-/
Some notes to have into account:

• in file1.php there are two forms: one to upload a file (example above) and another to save all the data into a database (that’s the reason I need the variable name).

• the example is just that, an example. I’m not really adding 5 to the variable requested, but I don’t want to copy and paste 100 lines of code to overwhelm everybody.

• the variable is also refreshed with javascript, So I see it there but I don’t really know how to assign a javascript variable to a php variable (if possible).

THANKS!!!

Answer 1

Here’s how I would do it.

The html:

<form id="form1" action="foo.php" method="post">
    <!-- form elements -->
</form>

<form id="form2" action="bar.php" method = "post">
    <input type="hidden" name="filename" value="" />
    <!-- other form elements -->
</form>

The javascript

$('#form1').submit(function(){
    var formdata = ''; //add the form data here
    $.ajax({
      url: "foo.php",
      type: "POST",
      data: formdata,
      success : function(filename){
            //php script returns filename
            //we apply this filename as the value for the hidden field in form2
            $('#form2 #filename').val(filename);
      }
    });
});

$('#form2').submit(function(){
    //another ajax request to submit the second form
   //when you are preparing the data, make sure you include the value of the field 'filename' as well  
   //the field 'filename' will have the actual filename returned by foo.php by this point
});

The PHP

foo.php

//receive file in foo.php

$filename = uniqid(); //i generally use uniqid() to generate unique filenames 
//do whatever with you file
//move it to a directory, store file info in a DB etc.

//return the filename to the AJAX request
echo $filename;

bar.php

//this script is called when the second form is submitted.
//here you can access the filename generated by the first form

$filename = $_POST['filename'];

//do your stuff here

use the Jquery Form plugin to upload the file via Ajax

$(document).ready(function(){
    $('yourform').submit(function(){        //the user has clicked on submit

        //do your error checking and form validation here

        if (!errors)
        {
            $('yourform').ajaxSubmit(function(data){        //submit the form using the form plugin
                alert(data);    //here data will be the filename returned by the first PHP script
            });
        }
    });
});

As you’ll notice, you haven’t specified either the POST data or the URL of the PHP script. ajaxSubmit picks up the POST data from the form automatically and submits it to the URL specified in the action of the form