如何在php中创建如下的多维数组？

Question

嗨，大家好，我正在尝试创建以下格式的多维数组，但我正努力做到这一点，我们将不胜感激。 谢谢。

    $movies = array();
    $sql = "SELECT u.username AS Username, group_concat(v.title) AS Video_title, group_concat(r.rate) AS Rating
    FROM `rating` AS r 
    JOIN `uploads` AS v ON r.video_id = v.vid_id 
    JOIN `users` as u ON u.user_id = r.rater_id 
    GROUP by Username" ;

    $query = mysqli_query($conn, $sql);

    while($result = mysqli_fetch_assoc($query)){
    $movies[] = $result;
    }

    foreach($movies as $k => $v){
    unset($movies[$k]);
    $movies[$v['Username']] = $v;
    }
    foreach($movies as $k => $v){
    unset($movies[$k]['Username']);
    }

    print_r($movies);

这是我目前得到的：

     [user1] => Array ( [Video_title] => Snakes on the Plane,
                                         Superman Returns,
                                         The Night Listner,
                                         Lady in the Water,
                                         Just my Luck,
                                         You me and Dupree 
                         [Rating] => 3.5,
                                     3.5, 
                                     3,
                                     2.5,
                                     3,
                                     2.5 
                     ) )

但我的目标是将其输出为以下格式：

 'user1' => 
          array('Lady in the Water' => 2.5,
                'Snakes on a Plane' => 3.5, 
                'Just My Luck' => 3.0, 
                'Superman Returns' => 3.5,
                'You, Me and Dupree' => 2.5, 
                'The Night Listener' => 3.0
               )

Answer 1

更改以下行：

$movies[$v['Username']] = $v;

至

$movies[$v['Username']][] = $v;

然后再试一次。 它将为每次迭代在“ Username索引下创建一个新数组。

Answer 2

尝试以下代码以获得目标结果：

    $movies = array();
    $sql = "

    SELECT u.username AS Username, v.title AS Video_title, r.rate AS Rating
    FROM `rating` AS r 
    JOIN `uploads` AS v ON r.video_id = v.vid_id 
    JOIN `users` as u ON u.user_id = r.rater_id 
    GROUP by Username" ;

    $query = mysqli_query($conn, $sql);

    while($result = mysqli_fetch_assoc($query)){
        $movies[$result['Username']][$result['Video_title']] = $result['Rating'];
    }

    print_r($movies);

Answer 3

你必须像下面这样做：

  $movies = array();
    $sql = "SELECT u.username AS Username, group_concat(v.title) AS Video_title, group_concat(r.rate) AS Rating
    FROM `rating` AS r 
    JOIN `uploads` AS v ON r.video_id = v.vid_id 
    JOIN `users` as u ON u.user_id = r.rater_id 
    GROUP by Username" ;
    $query = mysqli_query($conn, $sql);

    while($result = mysqli_fetch_assoc($query)){
        $movies[] = $result;
    }

    $new_movies = array(); // a new array variable

    foreach($movies as $k => $v){
        $array_keys = explode(',',$v['Video_title']); // explode Video_title string into array
        $trimmed_array_key=array_map('trim',$array_keys); // trim all spaces from each value of $trimmed_array_key array
        $array_values = explode(',',$v['Rating']); // explode Rating array
        $trimmed_array_values=array_map('trim',$array_values);//remove all spaces from $trimmed_array_values array
        $new_movies[$v['Username']] = array_combine($trimmed_array_key,$trimmed_array_values); // combine $trimmed_array_key array and $trimmed_array_values array and assign to corresponding user
    }


    echo "<pre/>";print_r($new_movies); // print new array

注意：-

这行代码：- $trimmed_array_key=array_map('trim',$array_keys);

将执行以下操作：

https://eval.in/766605

参考： -http : //php.net/manual/zh/function.array-combine.php