[英].val() doesn't work - can't find solution
$(document).ready(function() {
var login = $("#login").val();
var password = $('#password').val();
$('.login-button').click(function() {
alert(login);
});
});
HTML
<form method="post" class="login"> <p id="login-error"></p>
<p>
<label for="login">Username:</label>
<input type="text" name="login" id="login" placeholder="username">
</p>
<p>
<label for="password">Password:</label>
<input type="password" name="password" id="password" placeholder="password">
</p>
<p class="login-submit">
<button class="login-button">Login</button>
</p>
<p class="forgot-password">Fill your username and password.</p> </form>
什么也沒顯示! 怎么了 有什么辦法嗎? ID名稱正確
編輯:我還有另一個問題。 不想為此提出另一個問題。 我正在嘗試從validate.php獲得ajax答案(它在/views/admin/validate.php中-在打開domain.com/validate時運行)
$(document).ready(function() {
$('.login-button').click(function() {
var login = $("#login").val();
var password = $('#password').val();
$.ajax({
type: 'POST',
url: '/view/admin/validate.php',
data: {
login : login,
password : password
},
success: function(data){
$('#login-error').html(data);
}
});
});
});
validate.php
<?php
session_start();
$user = mysql_real_escape_string($_POST['login']);
$password = mysql_real_escape_string(sha1($_POST['password']));
$query = mysql_query("SELECT * FROM `users` WHERE name = '$user' AND pass = '$password' AND privileges = 'superuser'");
$num_rows = mysql_num_rows($query);
if($num_rows == '0') {
echo "Username and Password are incorrect! (Maybe you don't have permission!)";
}
elseif($num_rows == '1') {
$expire = time()*60*60*60*60;
setcookie("user","$user",$expire);
$_SESSION['user'] = $user;
include '/views/admin/admin.php';
}
?>
它應該返回用戶名和密碼不正確! (也許您沒有權限!)-但是沒有。任何解決方案??
var login = $("#login").val();
運行此行時,輸入為空。
用戶鍵入某些內容后,您需要獲取該值。
$('.login-button').click(function() {
var login = $('#login').val();
alert(login);
});
試試吧 :
$('.login-button').click(function() {
alert($("#login").val());
});
要么 :
$('.login-button').click(function() {
var login = $("#login").val();
alert(login);
});
要么 :
var login = "";
$('.login-button').click(function() {
login = $("#login").val();
alert(login);
});
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