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创建新注册时，我无法写入数据库....！我有既有的登录和注册部分，如下所示一个javascript ...这是更新的脚本和PHP脚本两者的代码版本的登录ANC使用jquery php mysql提交数据？
<!DOCTYPE html>
<html>
<head>
<title>Load </title>
<meta name="viewport" content="width=device-width, height=device-height, initial-scale=1.0"/>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.css" />
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.js"></script>
<script src="js/index.js"></script>
</head>
<body>
<div data-role="page" id="login" data-theme="b">
<div data-role="header" data-theme="a">
<h3>Login Page</h3>
</div>
<div data-role="content">
<form id="check-user" class="ui-body ui-body-a ui-corner-all" data-ajax="false">
<fieldset>
<div data-role="fieldcontain">
<label for="username">Enter your username:</label>
<input type="text" value="" name="username" id="username"/>
</div>
<div data-role="fieldcontain">
<label for="password">Enter your password:</label>
<input type="password" value="" name="password" id="password"/>
</div>
<input type="button" data-theme="b" name="submit" id="submit" value="Submit">
</fieldset>
<a href="#registerp" data-role="button">Register</a>
</form>
</div>
<div data-theme="a" data-role="footer" data-position="fixed">
</div>
</div>
<div data-role="page" id="registerp">
<div data-theme="a" data-role="header">
<h3>Register</h3>
</div>
<div data-role="content">
<form id="registerform" class="ui-body ui-body-a ui-corner-all" data-ajax="false">
<fieldset>
<div data-role="fieldcontain">
<label for="fname">First Name:</label>
<input type="text" value="" name="fname" id="fname"/>
</div>
<div data-role="fieldcontain">
<label for="lname">Last Name:</label>
<input type="text" value="" name="lname" id="lname"/>
</div>
<div data-role="fieldcontain">
<label for="uname">User Name:</label>
<input type="text" value="" name="uname" id="uname"/>
</div>
<div data-role="fieldcontain">
<label for="pwd">Enter your password:</label>
<input type="password" value="" name="pwd" id="pwd"/>
</div>
<div data-role="fieldcontain">
<label for="email">Email:</label>
<input type="text" value="" name="email" id="email"/>
</div>
<input type="button" data-theme="b" name="submit" id="register" value="Register">
</fieldset>
</form>
</div>
<div data-theme="a" data-role="footer" data-position="fixed">
<h3>Page footer</h3>
</div>
</div>
<div data-role="page" id="second">
<div data-theme="a" data-role="header">
<h3>Welcome Page</h3>
</div>
<div data-role="content">
Welcome
</div>
<div data-theme="a" data-role="footer" data-position="fixed">
<h3>Page footer</h3>
</div>
</div>
<script type="text/javascript">
$(document).on('pageinit', '#login', function(){
$(document).on('click', '#submit', function() { // catch the form's submit event
if($('#username').val().length > 0 && $('#password').val().length > 0){
// Send data to server through the ajax call
// action is functionality we want to call and outputJSON is our data
$.ajax({url: 'check.php',
data: "action=login&" + $('#check-user').serialize(),
type: 'post',
async: 'true',
dataType: 'json',
beforeSend: function() {
// This callback function will trigger before data is sent
$.mobile.showPageLoadingMsg(true); // This will show ajax spinner
},
complete: function() {
// This callback function will trigger on data sent/received complete
$.mobile.hidePageLoadingMsg(); // This will hide ajax spinner
},
success: function (result) {
if(result.status) {
$.mobile.changePage("#second");
} else {
alert('Log on unsuccessful!');
}
},
error: function (request,error) {
// This callback function will trigger on unsuccessful action
alert('Network error has occurred please try again!');
}
});
} else {
alert('Please fill all necessary fields');
}
return false; // cancel original event to prevent form submitting
});
});
</script>
<script type="text/javascript">
$(document).on('pageinit', '#registerp', function(){
$(document).on('click', '#register', function() {
if($('#uname').val().length > 0 && $('#pwd').val().length > 0){
// Send data to server through the ajax call
// action is functionality we want to call and outputJSON is our data
$.ajax({url: 'insert.php',
data: "action=register&" + $('#registerform').serialize(),
type: 'post',
async: 'true',
dataType: 'json',
beforeSend: function() {
// This callback function will trigger before data is sent
$.mobile.showPageLoadingMsg(true); // This will show ajax spinner
},
complete: function() {
// This callback function will trigger on data sent/received complete
$.mobile.hidePageLoadingMsg(); // This will hide ajax spinner
},
success: function (result) {
if(result.status) {
$.mobile.changePage("#second");
} else {
alert(' Try again later ! Server is busy !');
}
},
error: function (request,error) {
// This callback function will trigger on unsuccessful action
alert('Network error has occurred please try again!');
}
});
} else {
alert('Please fill all necessary fields');
}
return false; // cancel original event to prevent form submitting
});
});
</script>
</body>
</html>
While my PHP Script is simple as shown below... please help
<?php
$con=mysqli_connect("...............", "...........", ".........","........");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
$fname = mysqli_real_escape_string($con, $_POST['fname']);
$lname = mysqli_real_escape_string($con, $_POST['lname']);
$uname = mysqli_real_escape_string($con, $_POST['uname']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$password = mysqli_real_escape_string($con, $_POST['pwd']);
$action = $_POST['action'];
// Decode JSON object into readable PHP object
//$formData = json_decode($_POST['formData']);
$sql="INSERT INTO userdb (username, fname, lname, password, email) VALUES ('$uname', '$fname', '$lname', '$password','$email')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
if($action == 'register'){
$output = array('status' => true, 'message' => 'Registered');
}
echo json_encode($output);
?>
Insert php script doesnt work while the below register php script works fine.
<?php
// We don't need action for this tutorial, but in a complex code you need a way to determine Ajax action nature
$action = $_POST['action'];
// Decode JSON object into readable PHP object
//$formData = json_decode($_POST['formData']);
// Get username
$username = $_POST['username'];
// Get password
$password = $_POST['password'];
$db = @mysql_connect('..........', '........', '..........') or die("Could not connect database");
@mysql_select_db('users', $db) or die("Could not select database");
$result = mysql_query("SELECT `password` FROM `userdb` WHERE `username`= '$username'");
$r = mysql_fetch_assoc($result);
$pass_ret = $r['password'];
// Lets say everything is in order
if($action == 'login' && $password == $pass_ret){
$output = array('status' => true, 'message' => 'Login');
}
else
{
$output = array('status' => false, 'message' => 'No Login');
}
echo json_encode($output);
？>
*“它总是返回错误.....在我的注册脚本”* - 是？ –
请让我们知道错误 –
相同在我的代码的警报消息警报（'网络错误发生，请再试！'）; – user2939685