Einfügen von Daten in die Datenbank mit Jquery/Ajax & PHP

Question

Ich habe versucht, Daten an die Datenbank mit Ajax, aber ich bleibe ich stecken.

Ich nahm einfachen Code, um es zu testen, aber es hat nicht funktioniert, egal was ich mache.

HTML/Ajax-Code

<?php include("osb.php");?> 
<script type = "text/javascript" src = "http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> 

<!--we have our html form here where user information will be entered--> 
     <form action='osb.php' method='post' border='0' id="form1"> 
      <div id = "container"> <br> 
       <label>Name:   </label> <input type='text' id="name" name='name' /><br> <br> 
       <label>E-mail:   </label> <input type='text' id="email" name='email' /><br><br><br> 

       <input type='hidden' name='action' value='create' /> 
       <input type='button' value='Submit' id="submit" /> 
       <input type="reset" value="Reset" class="reset-org"> 
       <div> 

     </form> 

<script type = "text/javascript"> 
    $(function(){ 

     $('#submit').click(function(){ 


      $('#container').append('<img src = "img/ajax/ajax-loader.gif" alt="Currently loading" id = "loading" />'); 
      $.ajax({ 

       url: 'osb.php', 
       type: 'POST', 
       data: $('#form1').serialize(), 
       success: function(result){ 
        $('#response').remove(); 
        $('#container').append('<p id = "response">' + result + '</p>'); 
        $('#loading').fadeOut(500); 

       } 

      }); 

     }); 
    }); 
</script> 

PHP CODE

<?php 
//set connection variables 




$host = "localhost"; 
$username = ""; 
$password = ""; 
$db_name = ""; 

//connect to mysql server 
$mysqli = new mysqli($host, $username, $password, $db_name); 



//check if any connection error was encountered 
if(mysqli_connect_errno()) { 
    echo "Error: Could not connect to database."; 
    exit; 
} 

$action = isset($_POST['action']) ? $_POST['action'] : ""; 



if($action=='create'){ //the the user submitted the form 

    $data=$_POST['serialize']; 
    $name=$data['name']; //access data like this 
    $email=$data['email']; //access data like this 
//include database connection 

//our insert query query 
//$mysqli->real_escape_string() function helps us prevent attacks such as SQL injection 
    $query = "insert into 'user' VALUES ($name,$email)"; 
    mysqli_query($mysqli, $query); 

//execute the query 
    if($mysqli ->query($query)) { 
     //if saving success 
     echo "User was created."; 
    }else{ 
     //if unable to create new record 
     echo "Database Error: Unable to create record."; 
    } 
//close database connection 
    $mysqli->close(); 
} 



?> 

ich diesen Fehler jedes Mal stelle ich die Form:

Undefined index: serialize in C:\xampp\htdocs\php1\osb.php on line 29

Database Error: Unable to create record.

Answer 1

entfernen

$data=$_POST['serialize']; 
$name=$data['name']; //access data like this 
$email=$data['email']; //access data like this 

In

$name=$_POST['name']; 
$email=$_POST['email']; 

Grüße