通过PHP返回JSON响应

Question

如何返回以下响应？

成功时应返回：

{ status : "ok", data : 
  [ { franchisor_no : <franchisor number>
    , franchisor_status : uncollected | active | delivered | returned | exception
  , events_list : 
    [ { date: <date>, status : uncollected | active | delivered | returned | exception
       , description: <optional description>
       , code: <optional code, may map to the defined franchisor codes>
       , location: <optional location, such as city or hub>.
       ... raw_event: <the original event as received from the franchisor API. mandatory 
... } ] } .... ] }

我正在使用这段代码，但是没有向我的服务器发送响应。请告诉我这段代码中是否有任何错误？

<?php
$data = json_decode(file_get_contents("php://input"));

echo json_encode = [
    "status" => "ok",
    "data" => [
        [
            "franchisor_no" => "1210110080",
            "franchisor_status" => "exception",
            "events_list" => [
                [
                    "date" => "30-07-2023",
                    "status" => "exception",
                    "description" => "optional",
                    "code" => "optional",
                    "location" => "optional",
                    "raw_event" => "mandatory"
                ],
            ],
        ],
    ],
];

Answer 1

应该是这样

echo json_encode([
"status" => "ok",
"data" => [
    [
        "franchisor_no" => "1210110080",
        "franchisor_status" => "exception",
        "events_list" => [
            [
                "date" => "30-07-2023",
                "status" => "exception",
                "description" => "optional",
                "code" => "optional",
                "location" => "optional",
                "raw_event" => "mandatory"
            ],
        ],
    ],
]]);

您没有调用json_encode()函数。您只是在它前面加了一个等号。