ajax表单多个值提交不起作用

时间:2014-01-23 13:52:05

标签: php ajax forms jquery post

我尝试使用ajax发送数据。这是我的ajax代码。但我无法将数据发送到数据库。任何人都可以提供帮助吗?

$('#submit').click(function(){
        //get data to the variables
    var username = $("#username").val();
    var password = $("#password").val();
    var email = $("#email").val();
    var firstname = $("#firstname").val();
    var lasttname = $("#lastname").val();
    var age = $("#age").val();
    var gender = $("#gender").val();
    var country = $("#country").val();

    var dataString = 'username='+username + '&password='+ password + '&email='+ email + '&firstname='+firstname + '&age='+ age + '&gender='+gender + '&country='+ country;  
    //alert (dataString);return false;  
    $.ajax({  
        type: "POST",  
        url: "register-exec.php",  
        data: dataString,  
        success: function() {  
        $('#sendform').html("<div id='message'></div>");  
        $('#message').html("<h2>Contact Form Submitted!</h2>" + dataString)  
        .append("<p>We will be in touch soon.</p>")  
        .hide()  
        .fadeIn(1500, function() {  
        $('#message').append("<img id='checkmark' src='images/check.png' />");  

        });  
      }  
    }); 
    alert( "Data Saved: "); 

    });

});

我已经编写了数据库连接的代码,这是正确的。我检查了很多东西并在谷歌搜索,但我无法弄清楚是什么问题。 我的register-exec.php代码是..

<?php
    //Start session
    session_start();

    //Include database connection details
    require_once('config.php');


    //Connect to mysql server
    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {
        die('Failed to connect to server: ' . mysql_error());
    }

    //Select database
    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {
        die("Unable to select database");
    }


    //get the POST values
    $username = $_POST['username'];
    $password = $_POST['password'];
    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $email = $_POST['email'];
    $age = $_POST['age'];
    $gender = $_POST['gender'];
    $country = $_POST['country'];

    //Create INSERT query
    $qry = "INSERT INTO step-signin(username, password, firstname, lastname, email, age, gender, country) VALUES('$username','".md5($_POST['$password'])."','$firstname','$lastname','$email','$age','$gender','$country')";
    $result = @mysql_query($qry);

    //Check whether the query was successful or not
    if($result) {
        header("location: register-success.php");
        exit();
    }else {
        die("Query failed");
    }
?>

0 个答案:

没有答案