我尝试使用jquery插件实现表单验证。如果我单击提交按钮,除了执行php代码并创建cookie userid
和username
之外没有任何反应。添加$.ajax({
后,表单验证规则将被忽略,重定向失败,但没有$.ajax({
php代码将无法正常工作。验证,PHP代码和重定向如何一起工作?
我的HTML代码:
<head>
<meta charset="UTF-8" />
<title>
HTML Document Structure
</title>
<link rel="stylesheet" type="text/css" href="style.css" />
<link rel="stylesheet" href="themes/my-costum-theme.min.css" />
<link rel="stylesheet" href="themes/jquery.mobile.icons.min.css" />
<script src="http://code.jquery.com/jquery-2.1.3.min.js"></script>
<script src="http://code.jquery.com/mobile/1.4.5/jquery.mobile-1.4.5.min.js"></script>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.4.5/jquery.mobile.structure-1.4.5.min.css" />
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.14.0/jquery.validate.min.js"></script>
<!-- Einstellungen zur Defintion als WebApp -->
<meta name="apple-mobile-web-app-capable" content="yes" />
<meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">
<script src="loadagain.js"></script>
<script>
$(document).ready(function () {
$('#myform').validate({ // initialize the plugin
rules: {
username: {
required: true,
minlength: 2,
maxlength: 30
},
password: {
required: true,
minlength: 3,
maxlength: 30
}
},
submitHandler: function (form) { // for demo
$.ajax({
type: 'post',
url: 'loginprivate.php' //url where you want to post the stuff.
data:{
username: 'root',
password: ''
},
success: function(res){
//here you will get res as response from php page, either logged in or some error.
window.location.href = "http://localhost/loc/main.php";
}
});
return false; // for demo
}
});
});
</script>
</head>
<body>
<div class="ui-page" data-theme="b" data-role="page">
<div data-role="header"><h1>localcorps</h1></div>
<div id="wrapper1" style=" width: 90%; padding-right:5%; padding-left:5%" name="wrapper1">
<form name="login-form" id="myform" class="login-form" action="./loginprivate.php" method="post">
<div class="header1"></div>
<div class="content1">
<div data-role="fieldcontain">
<label for="username">Username:</label>
<input name="username" type="text" class="input username" id="username"/>
</div>
</div>
<div data-role="fieldcontain">
<label for="password">Password:</label>
<input name="password" type="password" class="input password" id="password"/>
</div>
<div class="footer">
<input type="submit" name="submit" value="Login" class="button"/>
</div>
</form>
</div>
</div>
</div>
</body>
我的PHP代码:
if(isset($_POST["submit"]))
{
$hostname='localhost';
$username='root';
$password='';
unset($_POST['password']);
$salt = '';
for ($i = 0; $i < 22; $i++) {
$salt .= substr('./ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789', mt_rand(0, 63), 1);
}
$_POST['password'] = crypt($_POST['password'],'$2a$10$'.$salt);
$new = 0;
try {
$dbh = new PDO("mysql:host=$hostname;dbname=search",$username,$password);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // <== add this line
$sql = "INSERT INTO users (username, password)
VALUES ('".$_POST["username"]."','".$_POST["password"]."')";
if ($dbh->query($sql)) {
echo "New Record Inserted Successfully";
}
else{
echo "Data not successfully Inserted.";
}
$new = $dbh->lastInsertId();
$dbh = null;
}
catch(PDOException $e)
{
echo $e->getMessage();
}
if ($new > 0)
{
$t = time() + 60 * 60 * 24 * 1000;
setcookie("username", $_POST['username'], $t);
setcookie("userid", $new , $t);
}
else
{
}
}
答案 0 :(得分:0)
所以,请使用ajax
validate()
来电
<script>
$(document).ready(function () {
$('#myform').validate({ // initialize the plugin
rules: {
username: {
required: true,
minlength: 2,
maxlength: 30
},
password: {
required: true,
minlength: 3,
maxlength: 30
}
},
submitHandler: function (form) {
// do other things for a valid form
form.submit(); //will submit your form.
return false; // disables default submit.
}
});
});
</script>
了解更多信息 HERE ,并特别关注示例2和3 。