在JSON结构中添加变量链接

Question

我想从JSON响应中添加链接:https://xxx.xxx/posters/poster.jpg，这里是我的代码：

/**
 * Get all movies
 *
 * @return \Illuminate\Http\Response
 */
public function getAllMovies()
{
    $movieQuery = DB::select('
                  SELECT
                  movies.m_poster AS poster,
                  movies.m_backdrop AS backdrop,
                  FROM movies
                  WHERE movies.show <> 0 AND movies.m_age <> "G"
                  GROUP BY movies.created_at,movies.m_id DESC
                  LIMIT 100');

    // Check if there is no movies
    if (empty($movieQuery)) {
        $movieQuery = null;
    }

下面是JSON响应：

        .....
        "poster": "04e33bbc8cdb8f09f6991acc7956dcae.jpg",
        "backdrop": "e3324cdb3cc03fe332e3cb233aae9c77.jpg",
        .....

如何让它变成这样：

              ......
                "poster": "https://xxx.xxx/posters/04e33bbc8cdb8f09f6991acc7956dcae.jpg",
                "backdrop": "https://xxx.xxx/backdrop/e3324cdb3cc03fe332e3cb233aae9c77.jpg",
             ......

Answer 1

如果您使用的是MySQL，那么可以使用CONCAT函数，如下所示：

$movieQuery = DB::select('
                  SELECT
                  CONCAT("https://xxx.xxx/posters/", movies.m_poster) AS poster,
                  CONCAT("https://xxx.xxx/backdrop/", movies.m_backdrop) AS backdrop,
                  FROM movies
                  WHERE movies.show <> 0 AND movies.m_age <> "G"
                  GROUP BY movies.created_at,movies.m_id DESC
                  LIMIT 100');

如果您想在显示到屏幕之前对它们进行迭代，上面Amir MB发布的其他方法也会起作用。

Answer 2

您可以对它们进行迭代,并将url添加到值的开头：

foreach ($json as $key => &$value) 
    $value = "https://xxx.xxx/$key/$value";