Javascript jQuery$.getJson调用未返回值

Javascript jQuery$.getJson调用未返回值,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我试图创建一个使用jQuery ajax调用$.getJson和jQuery模板插件动态创建信息的站点。在我的index.html页面中,我有以下一些代码: $(document).ready(function() { var jsonData ="string"; $.getJSON("load.php", function(data) { jsonData = data;

我试图创建一个使用jQuery ajax调用$.getJson和jQuery模板插件动态创建信息的站点。在我的index.html页面中,我有以下一些代码:

        $(document).ready(function() {
            var jsonData ="string";

            $.getJSON("load.php", function(data) {
                jsonData = data;
                console.log(jsonData);
            });

            console.log(JSON.stringify(jsonData));
            // Load template from our templates folder,
            // and populate the data from the post object.
            $("#test").loadTemplate('#template', jsonData);

        });
//extract user ratings and respective movies
$_SESSION['username'] = $username;
$user_query = mysqli_query($link, "SELECT * FROM Client WHERE Username = '$username'");
$user_row = mysqli_fetch_array($user_query);
$user_id = $user_row['ID'];

$query = mysqli_query($link, "SELECT * FROM Ratings INNER JOIN Movie ON Ratings.Movie_ID = Movie.MovieID WHERE Client_ID = $user_id ORDER BY Rating DESC");
//adding movies to array list
$result =  array();
while ($row = mysqli_fetch_array($query))
{
    $movie = $row['MovieURL'];
    $rating = $row['Rating'];
    array_push($result, array('moviePicture' => $movie, 'movieRating' => $rating));
}
//converting array list to json and echoing it
echo json_encode($result);
ajax调用执行load.php,其中包含以下一些代码:
//extract user ratings and respective movies
$_SESSION['username'] = $username;
$user_query = mysqli_query($link, "SELECT * FROM Client WHERE Username = '$username'");
$user_row = mysqli_fetch_array($user_query);
$user_id = $user_row['ID'];

$query = mysqli_query($link, "SELECT * FROM Ratings INNER JOIN Movie ON Ratings.Movie_ID = Movie.MovieID WHERE Client_ID = $user_id ORDER BY Rating DESC");
//adding movies to array list
$result =  array();
while ($row = mysqli_fetch_array($query))
{
    $movie = $row['MovieURL'];
    $rating = $row['Rating'];
    array_push($result, array('moviePicture' => $movie, 'movieRating' => $rating));
}
//converting array list to json and echoing it
echo json_encode($result);
我已经测试了$result,它确实创建了一个json对象,其中包含了它应该包含的所有数据。但是,当javascript在index.html页面中运行时,jsonData不会将其值更改为“string”(当我检查控制台日志时)。我认为问题在于函数(数据),因为它没有记录我在那里的控制台命令。任何帮助都将不胜感激。

正确的代码:

$(document).ready(function() {
    var jsonData ="string";

    $.getJSON("load.php", function(data) {
        jsonData = data;
        $("#test").loadTemplate('#template', jsonData);
    });
});
代码的问题是,在得到
load.php