如何返回一个有效的 JSON 对象？

Question

我刚刚进入 php，但我遇到了 JSON 对象返回的问题。

代码如下：

$stmt->bind_result($image_link, $start_time, $end_time, $viet_performer, $english_performer, $viet_event, $english_event, $day, $stage);

            while($stmt->fetch())
            {
                echo json_encode([["image_link" => $image_link,"start_time" => $start_time, "end_time" => $end_time, "viet_performer" => $viet_performer,
                                "english_performer" => $english_performer, "viet_event" => $viet_event, "english_event" => $english_event, "day" => $day,
                                "stage" => $stage]]);

                $stmt->bind_result($image_link, $start_time, $end_time, $viet_performer, $english_performer, $viet_event, $english_event, $day, $stage);
            }

这基本上输出：

{"image_link":"schedule_music.jpg","start_time":"17:00","end_time":"18:30","viet_performer":"","english_performer":"","viet_event ":"","english_event":"春节 音乐","day":0,"stage":1} {"image_link":"schedule_music.jpg","start_time":"11:00","end_time":"12:00","viet_performer":"","english_performer":"","viet_event":" Nh?c","english_event":"Music","day":1,"stage":0}

//再重复一次

但是，根据 JSON 验证器，它是一个无效的 JSON 对象。

我想像这样生成一个有效的 JSON：

[{"image_link":"schedule_music.jpg","start_time":"17:00","end_time":"18:30","viet_performer":"","english_performer":""," viet_event":"","english_event":"春节 音乐","day":0,"stage":1},{"image_link":"schedule_music.jpg","start_time":"11:00","end_time":"12:00","viet_performer" :"","english_performer":"","viet_event":"Nh?c","english_event":"Music","day":1,"stage":0}]

注意开始/结束括号以及逗号分隔符。

我怎样才能做到这一点？

这是一个更好的示例（在某处找到），说明我的输出看起来如何：

{
    "id": "a1",
    "session": "General",
    "name": "Exhibitor Setup Begins",
    "startTime": "0900",
    "details": "9am Exhibitor Hall",
    "png": "image",
    "speaker1": "Johnson",
    "speaker2": "Nelson",
    "speaker3": ""
}{
    "id": "b1",
    "session": "General",
    "name": "Conference Registration",
    "startTime": "1000",
    "details": "10am Noon Upper Level Lobby",
    "png": "image",
    "speaker1": "Jackson",
    "speaker2": "",
    "speaker3": ""
}

我希望输出是什么样的：

[
    {
        "id": "a1",
        "session": "General",
        "name": "Exhibitor Setup Begins",
        "startTime": "0900",
        "details": "9am Exhibitor Hall",
        "png": "image",
        "speaker1": "Johnson",
        "speaker2": "Nelson",
        "speaker3": ""
    },
    {
        "id": "b1",
        "session": "General",
        "name": "Conference Registration",
        "startTime": "1000",
        "details": "10am Noon Upper Level Lobby",
        "png": "image",
        "speaker1": "Jackson",
        "speaker2": "",
        "speaker3": ""
    }
]

Answer 1

关于您更新的代码，问题在于您是单独编码每个部分，而不是一次编码所有数据。试试这个：

$stmt->bind_result($image_link, $start_time, $end_time, $viet_performer, $english_performer, $viet_event,
    $english_event, $day, $stage);

$data = [];
while ($stmt->fetch()) {
    $data[] = [
        "image_link"        => $image_link,
        "start_time"        => $start_time,
        "end_time"          => $end_time,
        "viet_performer"    => $viet_performer,
        "english_performer" => $english_performer,
        "viet_event"        => $viet_event,
        "english_event"     => $english_event,
        "day"               => $day,
        "stage"             => $stage
    ];
}

echo json_encode($data);

---

基于早期版本的问题的回答：

看起来你想要一个对象数组：

echo json_encode([
    [
        "image_link"        => $image_link,
        "start_time"        => $start_time,
        "end_time"          => $end_time,
        "viet_performer"    => $viet_performer,
        "english_performer" => $english_performer,
        "viet_event"        => $viet_event,
        "english_event"     => $english_event,
        "day"               => $day,
        "stage"             => $stage
    ]
]);

请注意，包含字符串键的数组将被json_encode 转换为对象。在上面的代码中，内部数组变成了一个对象，而外部数组没有。

---

这可能是一种更直观的看待方式：

$objectOne = (object) [
    "image_link"        => $image_link,
    "start_time"        => $start_time,
    "end_time"          => $end_time,
    "viet_performer"    => $viet_performer,
    "english_performer" => $english_performer,
    "viet_event"        => $viet_event,
    "english_event"     => $english_event,
    "day"               => $day,
    "stage"             => $stage
];

$objectTwo = (object) [
    "image_link"        => $image_link,
    "start_time"        => $start_time,
    "end_time"          => $end_time,
    "viet_performer"    => $viet_performer,
    "english_performer" => $english_performer,
    "viet_event"        => $viet_event,
    "english_event"     => $english_event,
    "day"               => $day,
    "stage"             => $stage
];

echo json_encode([$objectOne, $objectTwo]);

Answer 2

正确的做法是先生成数据结构，然后对其进行编码：

$foo = [];
while($stmt->fetch()) {
    $foo []= [
         "image_link" => $image_link,
         "start_time" => $start_time,
         "end_time" => $end_time,
         "viet_performer" => $viet_performer,
         "english_performer" => $english_performer,
         "viet_event" => $viet_event,
         "english_event" => $english_event,
         "day" => $day,
         "stage" => $stage];
    ];
}
$json = json_encode($foo);