Question

我试图以这样的格式输出我的JSON：http://api.androidhive.info/json/movies.json并且我不确定如何通过我的PHP输出代码完成此操作。现在，它正在显示＆＃34; post＆＃34;在顶部，是一张地图，我不知道如何成功删除，在此处显示：http://shipstudent.com/complaint_desk/androidfriendsList.php?username=noah。如果您需要更多信息，请与我们联系。

PHP：

$rows = $stmt->fetchAll();


if ($rows) {

    $response["posts"]   = array();

    foreach ($rows as $row) 
    {
        $post = array();
        $post["username"] = $row["username"];
        $post["profile_picture"] = $row["profile_picture"];

        array_push($response["posts"], $post);
    }

    // echoing JSON response
    echo json_encode($response);


} else 
{   
    die(json_encode($response));
}

JSON：

{
    "posts": [
        {
            "username": "noah",
            "profile_picture": "https://shipstudent.com/animal/appphotos/978321177.jpg"
        },
        {
            "username": "e",
            "profile_picture": "https://shipstudent.com/complaint_desk/appphotos/owl.jpeg"
        },
    ]
}

所需格式：

[{
        "title": "Dawn of the Planet of the Apes",
        "image": "http://api.androidhive.info/json/movies/1.jpg",
        "rating": 8.3,
        "releaseYear": 2014,
        "genre": ["Action", "Drama", "Sci-Fi"]
    },
    {
        "title": "District 9",
        "image": "http://api.androidhive.info/json/movies/2.jpg",
        "rating": 8,
        "releaseYear": 2009,
        "genre": ["Action", "Sci-Fi", "Thriller"]
    },

Answer 1

$rows = $stmt->fetchAll();

$response = [];

foreach ($rows as $row) {
    $post = [
        "username" => $row["username"],
        "profile_picture" => $row["profile_picture"]
    ];
    $response[] = $post;
}

echo json_encode($response);

Answer 2

$response = array();

foreach ($rows as $row) 
{
    $post = array();
    $post["title"] => // define title;
    $post["image"] => // define image;
    $post["rating"] => // define rating;
    $post["releaseYear"] => // define releaseYear;
    $post["genre"] => // define genre, must be array();
    array_push($response, $post);
}

JSON输出数组

2 个答案: