太多的JSON信息

Question

我使用以下内容将数据从mysql转换为JSON：

$sql = "select img_name from user_gallery_images where user_id=$_SESSION[user_id]";

        $response = array();
        $posts = array();

        $result = $mysqli->query($sql);

        while($row = $result->fetch_array()){

            $file=$row['img_name']; 
            $fileDir = "gallery/$file.jpg";
            $posts[] = array('thumb'=> $fileDir, 'image'=> $fileDir);

        } 

        $response['posts'] = $posts;

        $fp = fopen('/home/public_html/users/'.$settings[username].'/gallery/gallery.json', 'w');
        $jsonData = stripslashes(json_encode($response));
        fwrite($fp, $jsonData);
        fclose($fp);

哪种方法效果很好并且创造了例如。

{"posts":
[
{"thumb":"gallery/tess1367386438.jpg","image":"gallery/tess1367386438.jpg"},
{"thumb":"gallery/tess1367386538.jpg","image":"gallery/tess1367386538.jpg"}
]
}

但是，我使用它的JQuery插件不会用外部“posts”容器读取它

问题：

如何剥离JSON中的外部“posts”容器以仅生成：

[
{"thumb":"gallery/tess1367386438.jpg","image":"gallery/tess1367386438.jpg"},
{"thumb":"gallery/tess1367386538.jpg","image":"gallery/tess1367386538.jpg"}
]

Answer 1

尝试

$jsonData = json_encode($response['posts']);