Question

我知道有很多类似的问题，但我找不到能解决问题的问题。

问题是我想检查我的表单中的用户/密码是否存在于mysql数据库中，我是通过调用表单的onsubmit属性中的javascript函数来实现的，该函数通过AJAX将输入发布到检查它的php和回显到javascript函数，该函数警告用户并返回false或返回true，以便表单提交。

HTML

<span id="error_message" style="color:#CC3300;font-size:15px" ></span><br/>
<form method="post" action="./includes/process_login.php" onsubmit="return validateForm()" name="login_form">
    <div>
        <label for="email">Email</label><br/>
        <input placeholder="Introduce tu email" type="text" id="email" name="email" maxlength="50" />
    </div>
    <div>
        <label for="password">Password</label><br/>
        <input placeholder="Introduce tu password" type="password" name="password" id="password" maxlength="16" />
    </div>
    <div>
        <a class="passwordReset" href="./admin/passwordRecovery.php">He olvidado mi contraseña...</a>
    </div>
    <input type="submit" id="submit_button" value="ENTRAR" />
</form>

JAVASCRIPT / AJAX

function validateForm() {
    var pattern = new RegExp(/* Email validation pattern */);
    var isValid;

    if (pattern.test($("#email").val())) {  
        $.ajax({
            url: "../includes/process_login.php",
            cache: false,
            type: "POST",
            data: "email=" + $("#email").val() + "&password=" + $("#password").val(),
            success: function(data){
                if (data == '1') {
                    isValid = true;
                } else {
                    $("#error_message").text('Email/Password incorrectos!CCC');
                    isValid = false;
                }
            },

            error: function(){
                $("#error_message").text('Email/Password incorrectos!');
                isValid = false;
            }
        });
} else {
    $("#error_message").text('El email debe tener un formato válido');
    isValid = false;

}

return isValid;
}

PHP验证和表单操作：

<?php
include_once 'db_connect.php';
include_once 'functions.php';

sec_session_start();

if (isset($_POST['email'], $_POST['password'])) {
    $email = $_POST['email'];
    $password = $_POST['password'];

    if (login($email, $password, $mysqli) == true) {
        // Login success 
    header("Location: ../graficas.php");
        echo '1';
    } else {
        // Login failed 
    echo '0';
    }
} else {
    // The correct POST variables were not sent to this page.
    echo '0';
}

exit;

当电子邮件输入符合电子邮件正则表达式时，表单始终提交，因此问题必须在ajax函数内。但我无法弄清楚它是什么，几个小时后，我比以往任何时候都更接近庇护。

提前致谢！

Answer 1

在validateForm()内添加此内容以阻止表单提交

e.preventDefault();

喜欢这个

function validateForm(e){
  var pattern = new RegExp(/* Email validation pattern */);
  var isValid;
  if (pattern.test($("#email").val())) {  
   e.preventDefault();
   //Do rest of your code

Answer 2

我终于可以通过将表单和所有逻辑放在一个PHP文件中来完成它：

    <!-- The HTML login form -->
    <form method="post" action="<?=$_SERVER['PHP_SELF']?>">
        <div>
            <label for="username">Login</label><br/>
            <input placeholder="Introduce tu login" type="text" id="user" name="username" maxlength="50" />
        </div>
        <div>
            <label for="password">Password</label><br/>
            <input placeholder="Introduce tu password" type="password" name="password" id="password" maxlength="16" />
        </div>
        <div><a class="passwordReset" href="./admin/passwordRecovery.php">He olvidado mi contraseña...</a></div>
        <input type="submit" id="submit_button" name="submit" value="ENTRAR" />
    </form>

    <?php
    if (isset($_POST['submit'])){
        include_once 'includes/db_connect.php';
        include_once 'includes/functions.php'; 
        sec_session_start();

        $username = $_POST['username'];
        $password = $_POST['password'];     

            if (login($username, $password, $mysqli) == true) {
                // Login success
                echo '<script type="text/javascript">window.location.replace("graficas.php");</script>';

            } else {
                // Login failed
                echo '<script type="text/javascript">document.getElementById(\'error_message\').innerHTML = "Login/Password incorrectos!";</script>';
            }
    }
    ?>

通过回声＆＃39;生成JS脚本当需要修改元素并避免AJAX问题时。

这里的关键元素是

isset($_POST['submit'])

检查表单是否先前已提交过。

我从这个很好的教程中提出了这个想法：http://w3epic.com/php-mysql-login-system-a-super-simple-tutorial/

onsubmit没有使用ajax mysql验证

2 个答案: