Question

我有一张邀请表。我从postman传递json格式的数据。

我想一次发送许多邀请。所以我想插入多个邀请。

我该怎么做？

我创建了一个邀请。

邀请：

obj.length

sendInvite.php

undefined

来自邮递员的输入：

    class Invitation
{
    private $sender_id,$date,$invitee_no,$status;

    function Invitation($sender_id,$date,$invitee_no,$status)
    {

        $this->sender_id = $sender_id;
        $this->date= $date;
        $this->invitee_no = $invitee_no;
        $this->status = $status;

    }
    function sendInvite()
    {

        $database = new Database(ContactsConstants::DBHOST,ContactsConstants::DBUSER,ContactsConstants::DBPASS,ContactsConstants::DBNAME);
        $dbConnection = $database->getDB();

        $stmt = $dbConnection->prepare("select * from Invitation where invitee_no =?");
        $stmt->execute(array($this->invitee_no));
        $rows = $stmt->rowCount();

        if($rows > 0)
        {
            $response = array("status"=>-3,"message"=>"Invitation exists.");
            return $response;
        }

            $stmt = $dbConnection->prepare("insert into Invitation(date,invitee_no,status) values(?,?,?)");
            $stmt->execute(array($this->date, $this->invitee_no, $this->status));
            $rows = $stmt->rowCount();
            $Id = $dbConnection->lastInsertId();

            $stmt = $dbConnection->prepare("select * from Invitation where sender_id=?");
            $stmt->execute(array($Id));
            $invitation = $stmt->fetchAll(PDO::FETCH_ASSOC);

            if ($rows < 1) {
                $response = array("status" => -1, "message" => "Failed to send Invitation., unknown reason");
                return $response;
            } else {
                $response = array("status" => 1, "message" => "Invitation sent.", "Invitation:" => $invitation);
                return $response;
            }

    }
}

输出：

    <?php

error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);
ini_set('display_errors', '1');

require 'Invitation.php';

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

if(empty($jsonText))
{
    $response = array("status"=>-2,"message"=>"Empty request");
    die(json_encode($response));
}

$json = json_decode($jsonText);

$date= $json -> date;
$invitee_no = $json -> invitee_no;
$status = $json -> status;

$invitation = new Invitation("",$date,$invitee_no,$status);
$response = $invitation->sendInvite();

echo(json_encode($response));

?>

编辑：

在Send Invite（）函数中：

{

"date" : "12/08/2016",
"invitee_no" : "5258",
"status" : "1"
}

在senInvite.php文件中：

   {
  "status": 1,
  "message": "Invitation sent.",
  "Invitation:": [
    {
      "sender_id": "29",
      "date": "12/08/2016",
      "invitee_no": "5259",
      "status": "1"
    }
  ]
}

var dump：

if ($rows < 1) {

        $response = array("status" => -1, "message" => "Failed to send Invitation., unknown reason");
        echo(json_encode($response));

    } else {
        $response = array("status" => 1, "message" => "Invitation sent.", "Invitation:" => $invitation);
        echo(json_encode($response));

    }

给出语法错误：未通过'S'

我想接受这个作为json数组并将所有记录插入表中。

有人可以帮忙吗？谢谢..

Answer 1

  <?php

   error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);
   ini_set('display_errors', '1');

   require 'Invitation.php';

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

   if(empty($jsonText))
   {
      $response = array("status"=>-2,"message"=>"Empty request");
      die(json_encode($response));
   }

$response = array();

   $json = json_decode($jsonText);
  foreach ($json as $jsn) {
   foreach($jsn as $j)
    {
      $date= $j -> date;
      $invitee_no = $j -> invitee_no;
      $status = $j -> status;
      $invitation = new Invitation("",$date,$invitee_no,$status);

       $response[] = $invitation->sendInvite();


    } 
 }
echo(json_encode($response));
 ?>

I have used foreach for array.

如何获取json数组并插入数据库。 PHP

1 个答案: