Question

我使用flightstates API来创建实时航班状态。他们在api中提供警报消息。他们说“To use the Alerts API, the developer must implement a web service that accepts an HTTP POST and parses an Alerts Message, then set up a rule specifying that service as the destination.”。我创建了这样的Web服务：

$data = json_decode(file_get_contents('php://input'), true);
$que = mysqli_query($con,"insert into firetable (datas) values ('$data')");

但它存储值“Array”。如何将数据存储在数据库中？

 "alert":{  
          "event":{  },
          "dataSource":"Airline",
          "dateTimeRecorded":"2017-12-05T08:36:24.957Z",
          "rule":{  },
          "flightStatus":{  
             "flightId":"942163187",
             "carrierFsCode":"9W",
             "flightNumber":"527",
             "departureAirportFsCode":"DXB",
             "arrivalAirportFsCode":"COK",
             "departureDate":{  
                "dateLocal":"2017-12-05T12:25:00.000",
                "dateUtc":"2017-12-05T08:25:00.000Z"
             },
             "arrivalDate":{  
                "dateLocal":"2017-12-05T17:50:00.000",
                "dateUtc":"2017-12-05T12:20:00.000Z"
             },...............
............................

像这样的API响应（当我将邮件电子邮件ID作为服务提供时，我在我的电子邮件中收到此响应）。我如何将'flightId'，'flightNumber'等存储到数据库中？

Answer 1

// remove json_decode 
    $data = json_decode(file_get_contents('php://input'), true);

到

$data = file_get_contents('php://input');

$que = mysqli_query($con,"insert into firetable (datas) values ('$data')");

数据将以JSON格式存储在DB中。

Answer 2

只需使用json_decode函数：

$obj = json_decode($json,true); // this is the string you are storing in the database
echo $obj['alert'][0]['flightStatus'][0]['flightId']; // here is 942163187

有关该功能的更多信息： enter link description here

在PHP POST Web服务中重用警报API

2 个答案: