更新mysqli prepared语句返回null json response PHP

时间:2016-10-29 18:57:47

标签: php json mysqli

在此Web服务中,用户数据已更新并返回JSON响应。每当我在浏览器中运行查询时,它都能完美运行。但是当我从Android应用程序访问webservice时,响应为空。我的代码是,

UpdateData.php(web服务)

<?php
require_once 'DB_Functions.php';
$db = new DB_Functions();

// receiving the post params
$name = "Amey";
$address = "PPR2";
$email = "2653@abc.com";
$contact = "123456";
$vid = "zh0000000";

$user=$db->updateUser($name, $address,  $email, $contact, $vid);
echo json_encode($user);
?>

DBFunction.php

<?php
class DB_Functions {
    private $conn;
    // constructor
    function __construct() {
        require_once 'DBConnect.php';
        // connecting to database
        $db = new DBConnect();
        $this->conn = $db->connect();
    }
    // destructor
    function __destruct() {

    }
    public function updateUser($name, $address, $email, $contact, $voter_id) {
        $sql = "UPDATE user SET user_name=?, user_address=?, user_email_id=?, user_mob_no=? WHERE voter_id=?";  
        $stmt =$this->conn->prepare($sql);
        $stmt->bind_param('sssss', $name, $address, $email, $contact, $voter_id);
        $stmt->execute();
        if ($stmt->errno) {
          $variable= "FAILURE!!! " . $stmt->error;
        }
        $variable= "Updated";
        $stmt->close();
        return $variable;
    }
}
?>

任何帮助将不胜感激。

0 个答案:

没有答案