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up vote 2 down vote favorite
1

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.

This is my json data.

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

This is my Php code.

<?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.

   <?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.

share|improve this question
1  
Also show us the content of db.php. – Amit Mar 20 '14 at 4:34
    
Your code looks good i.e. parsing JSON and insert query I suppose its related to connection and db select. Use mysql_error() after the query to see what happened. – Abhik Chakraborty Mar 20 '14 at 4:37
    
I think you have problem with the connection. As Amit said, show your db.php code – 4EACH Mar 20 '14 at 4:37
    
use error_reporting(E_ALL); – 4EACH Mar 20 '14 at 4:42
    
@amit have updated my db connection code in my question pls check it – user3427551 Mar 20 '14 at 4:42
up vote 5 down vote accepted 
 $json = file_get_contents('php://input');
 $obj = json_decode($json,true);

I think ur passing wrong variable. you should pass $json in json_decode as above..

share|improve this answer
up vote 2 down vote

There is no such variable as $data. Try

$obj = json_decode($json,true);

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

share|improve this answer
    
good catch !! even I tried his code locally but I had it correct while checking it so missed that completely that he was using wrong variable name !! – Abhik Chakraborty Mar 20 '14 at 4:40
up vote -1 down vote

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

<?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);
}

?>
share|improve this answer
    
What does this offer over the other two answers here? – Martijn Pieters♦ Nov 30 at 13:33
up vote -3 down vote 
$string=mysql_real_escape_string($json);
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