如何返回有效的 JSON 对象 - How to return a valid JSON object?

我刚刚进入php，我对JSON对象的返回有问题。

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"，"日"：0，"舞台"：1} {"image_link"："schedule_music.jpg"，"start_time"："11：00"，"end_time"："12：00"，"viet_performer"："，"english_performer"："，"viet_event"："Nh？c"，"english_event"："音乐"，"天"：1，"舞台"：0}
重复一次

但是，根据 JSON 验证器的说法，它是一个无效的 JSON 对象。

我想生成一个有效的 JSON，如下所示：

[{"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"，"天"：0，"舞台"：1}，{"image_link"："schedule_music.jpg"，"start_time"："11：00"，"end_time"："12：00"，"viet_performer"："："english_performer"："，"viet_event"："Nh？c"，"english_event"："音乐"，"天"：1，"舞台"：0}]

请注意开始/结束括号以及逗号分隔符。

我怎样才能做到这一点？

下面是一个更好的示例(在某处找到(，说明我的输出类似的样子：

{
    "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": ""
}

以及我希望输出看起来像这样：

[
    {
        "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": ""
    }
]

关于更新的代码，问题是您单独编码每个部分，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);

基于问题的早期版本的回答：

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通过 json_encode 转换为对象。在上面的代码中，内部数组成为对象，而外部数组则不是。

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]);

正确的方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);