Pass jquery - generated variable to php in same page - php

Question

I have a script that gets tha data on the row of a button when that button is clicked. The id of the button is id='show-button'. This is the script:
<script>
$(document).ready(function(){
$(".show-button").click(function() {
var $row = $(this).closest("tr"); // Find the row
var names = $row.find(".name").text(); // Find the name
var surname = $row.find(".surname").text(); // Find the surname
var lecturer_id = names."_".surname;
$("#show_dialog").dialog({autoOpen: false});
$(".show-button").on("click", function() {$("#show_dialog").dialog("open");});
});
});
</script>
The last two significant lines open a jquery dialog box.
With that, i mean these lines:
$("#show_dialog").dialog({autoOpen: false});
$(".show-button").on("click", function() {$("#show_dialog").dialog("open");});
Now, I need to pass the value of var lecturer_id to a php script outside this code, but inside the same document. This php code will generate the content of the dialog crated by these two lines. Lets assume that I just want to echo the variable passed inside the dialog box (with the php).
Any idea on how to make it work?

Answer 1

Your question is not 100% clear, but, just an idea, if I got you right:
<script>
$(document).ready(function(){
$(".show-button").click(function() {
var $row = $(this).closest("tr"); // Find the row
var names = $row.find(".name").text(); // Find the name
var surname = $row.find(".surname").text(); // Find the surname
var lecturer_id = names."_".surname;
$.post( "test.php", { names: names, surname: surname; lecturer_id: lecturer_id })
.done(function( data ) {
$("#show_dialog")[0].innerHTML = data ;
$("#show_dialog").dialog({autoOpen: false});
$(".show-button").on("click", function() {$("#show_dialog").dialog("open");});
});
});
});
</script>
And I agree with #JayBlanchard you don't even need any ajax call here, just generate your html like:
$(document).ready(function(){
$(".show-button").click(function() {
var $row = $(this).closest("tr"); // Find the row
var names = $row.find(".name").text(); // Find the name
var surname = $row.find(".surname").text(); // Find the surname
var lecturer_id = names."_".surname;
$("#show_dialog")[0].innerHTML = ' Name = '+names +'; Surname = '+surname ;
$("#show_dialog").dialog({autoOpen: false});
$(".show-button").on("click", function() {$("#show_dialog").dialog("open");});
});
});

Answer 2

You can use jQuery post or ajax.
$.post( "test.php", { name: "John", time: "2pm" })
.done(function( data ) {
alert( "Data Loaded: " + data );
});
test.php will be the reciving end of php, where it expects data sent by jquery.
{ name: "John", time: "2pm" } will be the data you are wishing to send off to php.
data will be the the data output by php.
refer to http://api.jquery.com/jquery.post/ for more information