在提交表单之前检查用户名是否可用
我想检查用户名是否在提交表单之前进行。在提交表单之前检查用户名是否可用
据我所知,我必须使用AJAX从JavaScript数据库中获取数据。我如何将用户名发送到PHP文件?
这是我的形式:
<form id="loginForm" action="register.php" method="post">
<p>Register:</p>
<p style="text-align: left;">Full name: <br><input type="text" name="name" required/></p>
<p style="text-align: left;">Email: <br><input type="text" name="email" required/></p>
//Username
<p style="text-align: left;">Username: <br><input id="username" type="text" name="username" onkeyup="validateUsername(value);" required/></p>
<span id="usernameError" style="display:none;border:1px solid red;">Username can only contain a-z, 0-9 and must be at least 6 characters loong</span>
<span id="usernameTaken" style="display:none;border:1px solid red;">Username taken</span>
<p style="text-align: left;">Password: <br><input type="password" name="password" required/></p>
<input type="submit" value="Register">
</form>
这是validateUsername()
功能:
function validateUsername(username) {
var re = /[a-zA-Z0-9]/;
alert(username.length);
if(re.test(username) && username.length > 5) {
document.getElementById('username').style.backgroundColor = "green";
document.getElementById('usernameError').style.display= "none";
--error;
} else {
document.getElementById('username').style.backgroundColor = "red";
document.getElementById('usernameError').style.display= "block";
++error;
}
//here i want to check if the user name is taken
}
如果用户名称被占用,我想显示'usernameTaken'跨度。 否则,我想隐藏它。
这里是PHP文件来检查,如果用户名是已经在数据库:
<?php
session_start();
define('DB_NAME', 'madsanker_dk_db');
define('DB_USER', 'madsanker_dk');
define('DB_PASSWORD', 'MyPassword');
define('DB_HOST', 'mysql43.unoeuro.com');
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' .mysqli_error());
}
$db_selected = mysqli_select_db($link, DB_NAME);
if (!$db_selected) {
die('Could not connect: ' .mysqli_connect_error());
}
$username = //The username;
$username = mysqli_real_escape_string($link,$username);
$sql = "SELECT * FROM mainLogin WHERE username = '$username'";
$result = mysqli_query($link, $sql);
$count = mysqli_num_rows($result);
if($count == 0) {
//if the username is NOT taken
return true;
} else {
//if the username IS taken
return false;
}
mysqli_close($link);
?>
这是如何完成的?
JS - JQUERY AJAX
$.ajax({
url: 'register.php', data: {action: 'isUserNameTaken', params: [username]},
type: 'post',
success: function(data) {
//Do Something
}
});
PHP
<?php
function isUserNameTaken($username) {
//Do Something;
}
if(!empty($_POST['action'])) {
$action = $_POST['action'];
switch($action) {
case 'isUserNameTaken':
$username = '';
if(!empty($_POST['params'])) {
$username = $_POST['params'][0];
}
isUserNameTaken($username);
break;
}
}
?>
'isset &&!empty'是无意义的,只需使用'!empty'。 – deceze
@deceze感谢您的评论,我编辑了我的答案,虽然说实话,我不是一个PHP开发人员,我只是发现问题做了一个快速的研究,并且想分享我想出来的结果:) –
你可以这样来做: 在客户端:
function validateUsername(){
if(//test username input for length...) {
$.ajax({
type: 'POST',
url: 'validate.php',
data: { username: username },
success: function(response) {
if(response==0){
//username is valid
}
elseif(response==1){
//username is already taken
}
elseif(response==2){
//connection failed
}
}
});
}
else{
//display "username is too short" error
}
}
validate.php:
<?php
session_start();
define('DB_NAME', 'madsanker_dk_db');
define('DB_USER', 'madsanker_dk');
define('DB_PASSWORD', 'MyPassword');
define('DB_HOST', 'mysql43.unoeuro.com');
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' .mysqli_error());
echo json_encode(2);
}
else{
$db_selected = mysqli_select_db($link, DB_NAME);
if (!$db_selected) {
die('Could not connect: ' .mysqli_connect_error());
echo json_encode(2);
}
else{
$username = $_POST["username"];
$username = mysqli_real_escape_string($link,$username);
$sql = "SELECT * FROM mainLogin WHERE username = '$username'";
$result = mysqli_query($link, $sql);
$count = mysqli_num_rows($result);
if($count == 0) {
//if the username is NOT taken
echo json_encode(0);
}else {
//if the username IS taken
echo json_encode(1);
}
mysqli_close($link);
}
}
?>
您还可以通过jQuery中与的onkeyup函数调用它,以便使该功能validateUsername每当用户输入东西到输入字段,用户名将被检查....
U可以使用AJAX使用元素的ID或jQuery的 – devpro
哪里是你的问题在此代码发送用户名到PHP?您是否测试过该用户名正确地出现在ajax请求中? –
PHP应该返回某种文档,它不是函数返回。最简单的例子是只打印“true”或“false”。 ** a ** -jax调用是** a ** - 异步调用,所以在验证函数中禁用提交按钮,然后在来自ajax的回调表明所有内容都很好时重新启用它。 'if(this.response ===“true”);'(或经过一些合理的超时,例如2秒) –