Question

我刚刚进入php，我遇到了返回JSON对象的问题。

以下是代码：

$stmt->bind_result($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：

{＆＃34;链接＆＃34;：＆＃34; schedule_music.jpg＆＃34;＆＃34; START_TIME＆＃34;：＆＃34; 17：00＆＃34;＆＃34; END_TIME＆＃ 34;：＆＃34; 18：30＆＃34;＆＃34; viet_performer＆＃34;：＆＃34;＆＃34;＆＃34; english_performer＆＃34;：＆＃34;＆＃34 ;, ＆＃34; viet_event＆＃34;：＆＃34;＆＃34;＆＃34; english_event＆＃34;：＆＃34;四环素   音乐＆＃34;＆＃34;天＆＃34;：0，＆＃34;级＆＃34;：1}   {＆＃34;链接＆＃34;：＆＃34; schedule_music.jpg＆＃34;＆＃34; START_TIME＆＃34;：＆＃34; 11：00＆＃34;＆＃34; END_TIME＆＃34 ;: ＆＃34; 12：00＆＃34;＆＃34; viet_performer＆＃34;：＆＃34;＆＃34;＆＃34; english_performer＆＃34;：＆＃34;＆＃34;＆＃34 ; viet_event＆＃34;：＆＃34;新罕布什尔C＆＃34;＆＃34; english_event＆＃34;：＆＃34;音乐＆＃34;＆＃34;天＆＃34;：1，＆＃34;阶段＆＃34;：0}

//再次重复

但是，根据JSON验证器，它是一个无效的JSON对象。

我想生成一个有效的JSON：

[{＆＃34;链接＆＃34;：＆＃34; schedule_music.jpg＆＃34;＆＃34; START_TIME＆＃34;：＆＃34; 17：00＆＃34;＆＃34; END_TIME＆＃34;：＆＃34; 18：30＆＃34;＆＃34; viet_performer＆＃34;：＆＃34;＆＃34;＆＃34; english_performer＆＃34;：＆＃34;＆＃34; ＆＃34; viet_event＆＃34;：＆＃34;＆＃34;＆＃34; english_event＆＃34;：＆＃34;四环素音乐＆＃34;＆＃34;天＆＃34;：0，＆＃34;级＆＃34;：1}，{＆＃34;链接＆＃34;：＆＃34; schedule_music.jpg＆＃34;，＆＃34; START_TIME＆＃34;：＆＃34; 11：00＆＃34;＆＃34; END_TIME＆＃34;：＆＃34; 12：00＆＃34;＆＃34; viet_performer＆＃34;：＆＃ 34;＆＃34;＆＃34; english_performer＆＃34;：＆＃34;＆＃34;＆＃34; viet_event＆＃34;：＆＃34;新罕布什尔C＆＃34;＆＃34; english_event＆＃34;：＆＃34;音乐＆＃34;＆＃34;天＆＃34;：1，＆＃34;级＆＃34;：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？

这是一个更好的例子（在某处找到）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": ""
}

我希望输出看起来像：

[
    {
        "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);

基于早期版本问题的回复：

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由json_encode转换为对象。在上面的代码中，内部数组变为对象，而外部数组则不是。

这可能是一种更直观的方式来看待它：

$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

正确的方法是首先生成数据结构，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);

如何返回有效的JSON对象？

2 个答案: