Question

I am struggling with the post() method.

I’ve been reading several posts on here and the jquery forums, and just trying to get a simple piece of code working is proving difficult. My ultimate goal is to pass a div #exportData and all it’s children to test.php.

I started with:

$.post("ajax/test.php", $("body").html());

To my mind this should return the entire contents of the current page to test.php (unless test.php requires a holding div or element to receive the content). Currently it only returns the blank page.

I then tried looking at the parameters for post() if I need to manipulate these:

$.ajax({
        type: 'POST',
        url: ajax/test.php,
        data: data,
        success: success,
        dataType: dataType
    });

Also declared a variable:

var data = { 
        html: #exportData
    };

That bit failed of course. I don’t know how to define the data in the parameters, or if this is the right place to do it.

Although I would have thought if:

 $.post("ajax/test.php", $("body").html());

would have worked then presumably I can substitute “body” for any class, id or selector I like.

Also does the submit button need certain parameters to tie it to the post function. At the moment it is purely:

<input type="submit" id="submit" value="send" name="submit">

Sorry this is such a basic question.

Answer 1

You could do

var html = $("body").html();
var data = { 
        html: html
    };
$.post("ajax/test.php", data);

as the second parameter of $.post() is an object wich contains the data you want to send to the server.

To send the data you could do:

<input type="submit" id="submit" value="send" name="submit">

js

$('input#submit').click(function(e){
  //prevent submitting the form (if there is a form)
  e.preventDefault();
    var html = $("body").html();
    var data = { 
            html: html
        };
    $.post("ajax/test.php", data);
});

server side you receive the data

 $html = $_POST['html']