小编Tra*_*eld的帖子

如何检查表单中的任何字段是否为空

每当我在表单中提交某些内容时,我想检查是否有任何字段为空.到目前为止,我所拥有的不起作用

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$username = $_POST['username'];
$password = $_POST['password'];
$passwordconf = $_POST['passwordconf'];
$email = $_POST['email'];
$securityq = $_POST['securityq'];
$qanswer = $_POST['qanswer'];

if(empty($firstname) || empty($lastname) || empty($username) || empty($password) || empty($passwordconf) || empty($email) || empty($securityq) || empty($qanswer))
{
    echo "You did not fill out the required fields.";
}
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和形式

<form name="registrationform" action="register.php">
    First Name:<input type="text" name="firstname">
    Last Name:<input type="text" name="lastname">
    Email:<input type="text" name="email">
    Username:<input type="text" name="username">
    Password:<input type="password" name="password">
    Confirm Password:<input type="password" name="passwordconf">
    Security Question:<input type="text" name="securityq"> …
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html php forms

8
推荐指数
1
解决办法
9万
查看次数

如何检查表单中的任何字段是否为空

我正在为我的网站构建一个登录系统,我需要一种方法来检查表单中的所有字段是否都填写完毕.我也遇到错误消息的问题,说我的电子邮件格式不正确.

<?php
$con = mysql_connect(localhost, 262096, 9201999);
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}
mysql_select_db("262096", $con);
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$username = $_POST['username'];
$password = $_POST['password'];
$passwordconf = $_POST['passwordconf'];
$email = $_POST['email'];
$securityq = $_POST['securityq'];
$qanswer = $_POST['qanswer'];

if(empty($firstname) || empty($lastname) || empty($username) || empty($password) ||
    empty($passwordconf) || empty($email) || empty($securityq) || empty($qanswer))
{
    echo "You did not fill out the required fields.";
}

$uname = "SELECT * FROM users WHERE username='{$username}'";
$unamequery = mysql_query($uname) …
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html php forms html5

6
推荐指数
1
解决办法
5万
查看次数

无法在 XAMPP 中启动 MySQL?

所以我试图使用我制作的这个登录系统,它开始给我一些奇怪的错误。我发现这是因为 MySQL 没有启动。所以我试图启动它,但它不会。这是日志:

10:06:12 PM [mysql] Attempting to start MySQL app...
10:06:12 PM [mysql] Status change detected: running
10:06:13 PM [mysql] Status change detected: stopped
10:06:13 PM [mysql] Error: MySQL shutdown unexpectedly.
10:06:13 PM [mysql] This may be due to a blocked port, missing dependencies, 
10:06:13 PM [mysql] improper privileges, a crash, or a shutdown by another method
10:06:13 PM [mysql] Check the "/xampp/mysql/data/mysql_error.log" file
10:06:13 PM [mysql] and the Windows Event Viewer for more clues
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以下是错误日志 (mysql_error.log) 中一些更值得注意的行:

130318 …
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mysql xampp

2
推荐指数
1
解决办法
6万
查看次数

调用未定义的方法 PDOStatement::bind_result()?

我不断收到此错误消息,但不知道为什么。这是一个登录脚本,我在这里复制了表单因为我不知道如何使用 PHPASS 进行登录。这是我的代码:

$email      = ($_POST['email']);
$pass       = ($_POST['pass']);

require 'connect.php';
require 'PasswordHash.php';
$hash = '*';
    $login = $con->prepare("SELECT password FROM basicuserinfo WHERE email=:email");
    $login->bindParam(':email', $email);
    $login->execute();
    $login->bind_result($hash);
    if (!$login->fetch() && $con->errno)
        die();
    if ($hasher->CheckPassword($pass, $hash)) {
        $what = 'Authentication succeeded';
    } else {
        $what = 'Authentication failed';
    }
    unset($hasher);
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php pdo phpass

1
推荐指数
1
解决办法
3965
查看次数

如何检查电子邮件是否已经注册?

我正在尝试检查是否已使用PDO注册了电子邮件,但是到目前为止,我还没有找到一种很好的方法来进行此操作。到目前为止,这就是我所拥有的。如果有帮助,则连接变量为“ $ con”。另外,我是否需要在同一文档中使用连接变量,还是在包含的文档中使用它?

<link rel="stylesheet" type="text/css" href="css\errors.css">
<?php
require 'header.php';
if (!$_POST['submit']) {
?>
<html>
<a href="register.html">Regsiter here!</a>
<?php
} else {
    require 'connect.php';
    $firstname = ($_POST['firstname']);
    $lastname = ($_POST['lastname']);
    $email = ($_POST['email']);
    $password = ($_POST['password']);
    $passwordconf = ($_POST['passwordconf']);

    $errorfields = "<p class='errormsg'>Please fill out all the fields!</p>";
    if (empty($firstname) || empty($lastname) || empty($email) || empty($password) || empty($passwordconf)) {
        echo "$errorfields";
    }
    $erroremail = "<p class='errormsg'>Email is not in name@domain format!</p>";
    $regex = "/^[a-z0-9]+([_.-][a-z0-9]+)*@([a-z0-9]+([.-][a-z0-9]+)*)+.[a-z]{2,}$/i";
    if (!preg_match($regex, $email)) {
        echo "$erroremail"; …
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php mysql database pdo

0
推荐指数
1
解决办法
5635
查看次数

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php ×4

forms ×2

html ×2

mysql ×2

pdo ×2

database ×1

html5 ×1

phpass ×1

xampp ×1