Sending multiple PHP files with AJAX

Question

Usually when I want to create new content in a php file and implement it somewhere in my site I do something like this (JS and PHP):

    case 'someCasename':
    ...
    include_once 'file.php';
    break;

...
success: function(data){
    $("#idSelector").html(data);
}

and the new content is displayed at the selected element.

But im curious on if its possible to send multiple php files at the same time. Not sure if this is really necessary and I've not come across a single time where I need to do this but I thought of the it and wanted to ask. This last code is most likely very very wrong but I just want to showcase what i mean:

    case 'someCasename':
...
$phpArray = array("file1" => include_once 'file1.php', "file2" => include_once 'file2.php', "file3" => include_once 'file3.php'):
echo $phpArray;
break;

...
success: function(data){
    $("#idSelector").html(data.file1);
    $("#idSelector").html(data.file2);
    $("#idSelector").html(data.file3);
}

Is something like this even remotly possible? You cant json_encode php files can you?

Check out jQuery's load() method. api.jquery.com/load

Jay Blanchard
– Jay Blanchard

2012-12-03 15:46:13 +00:00
Commented Dec 3, 2012 at 15:46 — Jay Blanchard
– Jay Blanchard, Commented Dec 3, 2012 at 15:46

jeroen · Accepted Answer · 2012-12-03 16:12:14Z

1

If I understand you correctly, your included php file echo out something that you want to use in the success function of the ajax call.

What you could do, is loop through all your includes and use output buffering to capture the output of these files in an array.

Something like:

case 'someCasename':
  ...
  $results = array();
  while (files_to_include)
  {
    ob_start;    // start output buffering
    include_once "your_file";
    $results[] = ob_get_clean();    // get the contents of the output buffer
  }
  echo json_encode($results);    // send all collected output as json to be used on the js side
  break;

Although this would work, it would of course be better to use functions that return values in your includes instead.

answered Dec 3, 2012 at 16:12

jeroen

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2 Comments

user1683645 Over a year ago

okay so on the other end I would retrieve the files like this: success: function(data){ var imported = $.parseJSON(data); $("#idSelector").html(imported.file1); $("#idSelector").html(imported.file2); $("#idSelector").html(imported.file3); }

jeroen Over a year ago

@user1683645 That's the general idea.

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