如何将json数组插入mysql数据库

Hi I'm trying to insert the json array into my MySQL database. I'm passing the data form my iphone there i have converted the data into json format and I'm passing the data to my server using the url its not inserting into my server.

嗨,我正在尝试将json数组插入我的MySQL数据库。我正在通过我的iphone传递数据,我将数据转换为json格式,并且我使用url将数据传递到我的服务器,而不是插入我的服务器。

This is my json data.

这是我的json数据。

[{"name":"0","phone":"dsf","city":"sdfsdf","email":"dsf"},{"name":"13123123","phone":"sdfsdfdsfsd","city":"sdfsf","email":"13123123"}]

This is my Php code.

这是我的Php代码。

<?php 

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

 //Database Connection
require_once 'db.php';

 /* insert data into DB */
    foreach($obj as $item) {
       mysql_query("INSERT INTO `database name`.`table name` (name, phone, city, email) 
       VALUES ('".$item['name']."', '".$item['phone']."', '".$item['city']."', '".$item['email']."')");

     }
  //database connection close
    mysql_close($con);

   //}
   ?>

My database connection code.

我的数据库连接代码。

   <?php

       //ENTER YOUR DATABASE CONNECTION INFO BELOW:
         $hostname="localhost";
         $database="dbname";
         $username="username";
         $password="password";

   //DO NOT EDIT BELOW THIS LINE
     $link = mysql_connect($hostname, $username, $password);
     mysql_select_db($database) or die('Could not select database');
 ?>

Please tell where I'm doing wrong in the above code basically I'm not a php developer I'm mobile application developer so I'm using the php as a server side scripting please tell me how to resolve this problem.

请告诉我在上面的代码中哪里做错了基本上我不是php开发人员我是移动应用程序开发人员所以我使用php作为服务器端脚本请告诉我如何解决这个问题。

4 个解决方案

#1

 $json = file_get_contents('php://input');
 $obj = json_decode($json,true);

I think you are passing the wrong variable. You should pass $json in json_decode as shown above.

我认为你传递了错误的变量。您应该在json_decode中传递$ json,如上所示。

#2

There is no such variable as $data. Try

没有像$ data这样的变量。尝试

$obj = json_decode($json,true);

Rest looks fine. If the error still persists, enable error_reporting.

休息看起来很好。如果错误仍然存在,请启用error_reporting。

#3

You are missing JSON source file. Create a JSON file then assign it to var data:

您缺少JSON源文件。创建一个JSON文件,然后将其分配给var数据:

<?php

require_once('dbconnect.php');

// reading json file
$json = file_get_contents('userdata.json');

//converting json object to php associative array
$data = json_decode($json, true);

// processing the array of objects
foreach ($data as $user) {
    $firstname = $user['firstname'];
    $lastname = $user['lastname'];
    $gender = $user['firstname'];
    $username = $user['username'];

    // preparing statement for insert query
    $st = mysqli_prepare($connection, 'INSERT INTO users(firstname, lastname, gender, username) VALUES (?, ?, ?, ?)');

    // bind variables to insert query params
    mysqli_stmt_bind_param($st, 'ssss', $firstname, $lastname, $gender, $username);

    // executing insert query
    mysqli_stmt_execute($st);
}

?>

#4

-3

$string=mysql_real_escape_string($json);

#1

 $json = file_get_contents('php://input');
 $obj = json_decode($json,true);