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透過PHP返回JSON響應

<p>如何回傳以下回應? </p><p>成功時應回傳:</p><p><br /></p> <pre class="brush:php;toolbar:false;">{ 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 ... } ] } .... ] }</pre> <p>我正在使用這段程式碼,但是沒有向我的伺服器發送回應。請告訴我這段程式碼中是否有任何錯誤? </p> <pre class="brush:php;toolbar:false;"><?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" ], ], ], ], ];</pre> <p><br /></p>
P粉988025835P粉988025835482 天前452

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  • P粉588660399

    P粉5886603992023-07-31 12:02:46

    應該是這樣

    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"
                ],
            ],
        ],
    ]]);

    您沒有呼叫json_encode()函數。您只是在它前面加了一個等號。

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