使用php从JSON数组填充web数据

我已经用JSON玩了几天了，我真的认为这是一种很酷的交换数据的方式。我正在使用jquery手机构建一个应用程序，我试图填充Json数据，到目前为止，我已经尝试了这种方法:

来自一个名为movie-details的json文件。我有这个:

{"movies":[{"id":"1","name":"Dabangg2","picUrl":"http:'/'/www.naz8.com'/images'/Dabangg2.jpg"},{"id":"2","name":"Talassh","picUrl":"http:'/'/www.naz8.com'/images'/talassh.jpg"},{"id":"3","name":"JAB TAK HAI JAAN","picUrl":"http:'/'/www.naz8.com'/images'/jthj.jpg"},{"id":"4","name":"Khiladi 786","picUrl":"http:'/'/www.naz8.com'/images'/khiladi786.jpg"}]}

，我可以通过以下方式获取数据来动态创建详细的listview:

<script type="text/javascript">
 $.getJSON("movie-details.json", function(movies){
   //Start off with an empty list every time to get the latest from server
   $('#movieList').empty();
   //add the movie items as list
   $.each(movies, function(i, movie){
     $('#movieList').append(generateMovieLink(movie));
   });
   //refresh the list view to show the latest changes
   $('#movieList').listview('refresh');
 });
  //creates a movie link list item
 function generateMovieLink(movie){
  //debugger;
  return '<li><a href="javascript:void(0)'
        + '" onclick="goToMovieDetailPage('''
        + movie.name 
        + ''','''
        + movie.picUrl +''')">' 
        + movie.name 
        + '</a></li>';
 }
 function goToMovieDetailPage(movieName, moviePicUrl){
  //create the page html template
  var moviePage = $("<div data-role='page' data-url=dummyUrl><div data-role='header' data-add-back-btn='true'><h1>"
                  + movieName + "</h1></div><div data-role='content'><img border='0' src='" 
                  + moviePicUrl + "' width=204 height=288></img></div><div data-role='footer' data-position='fixed'><h4>" 
                  + movieName + "</h4></div></div>");
  //append the new page to the page container
  moviePage.appendTo( $.mobile.pageContainer );
  //go to the newly created page
  $.mobile.changePage( moviePage );
 }  
</script>

如何填充来自php文件的数据，例如movie-details.php:

<?php
include('connection.php');
$var = array();
$sql = "SELECT * FROM movies";
$result = mysqli_query($con, $sql);
while($obj = mysqli_fetch_object($result)) {
$var[] = $obj;
}
echo '{"movies":'.json_encode($var).'}';
?>

我必须声明哪些变量来从php文件中的对象中获取json数据?

//只是一个想法…我很困惑……

var url ="....";

.getJSON美元(url,函数(…){

. each美元(电影。电影、函数(我,……)