Question

我的验证脚本使用php时出现问题;当用户只填写用户名表格并清空密码时，它仍然记录用户在其中应该向用户显示密码字段为空白错误。我是一个新的PHP，我希望你能帮助我。谢谢！

这是我检查登录的代码

<?php
$usernameErr = $passwordErr = "";
$username = $password = "";

if ($_SERVER["REQUEST_METHOD"] == "POST")
{

   if (empty($_POST['username']))
     {$usernameErr = "Username is required.";}
   else
     {$username =($_POST['username']);}

   if (empty($_POST['password']))
     {$passwordErr = "Password is required.";}
   else
     {$password =($_POST['password']);}
}
?>

<body>
<div id="header" align="center">
<h1>PT. Sumber Urip Alfindo</h1>
</div>
<br/>
<div id="content" align="center">
<form id="login" name="login" method="post" action="checklogin.php">
<table>
<tr>
<td>Username</td>
<td></td>
<td><input name="username" type="text" id="username"><span class="error"><?php echo $usernameErr;?></span></td>
</tr>
<tr>
<td>Password</td>
<td></td>
<td><input name="password" type="password" id="password"><span class="error"><?php echo $passwordErr;?></span></td>
</tr>
<tr>
<td colspan="3" align="center"><input type="submit" name="submit" value="Login"></td>
</tr>
</table>
</form>

<?php
$sql="SELECT * FROM $tbl_name WHERE usrname='$username'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);

if($count==1 && $username="admin")
{
header("location:mainadmin.php");
}

else if($count==1)
{
header("location:main.php");
}

else
{
echo "Wrong username or password";
}
?>

Answer 1

在任何人呻吟之前，我不会用mysqli / PDO替换mysql来回答这个问题。是的，使用它是错误的，但它与问题无关。

正确的模型：如果（没有错误）{记录该人}否则{做别的事}。

您的型号：检查错误。无论如何都要记录用户。

这就是你现在正在做的事情

// checking stuff
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
   if (empty($_POST['username']))
     {$usernameErr = "Username is required.";}
   // blah blah check check check 
}
// don't bother considering the error, just log them in anyway
$sql="SELECT * FROM $tbl_name WHERE usrname='$username'";
// etc

但你需要做的是：

// check for errors and store them

$errors=array(); // create an empty array to store errors
if (empty($_POST['username'])){
    $errors['usernameErr'] = "Username is required."; // add an error
}else{
    $username =($_POST['username']);
}

if (empty($_POST['password'])){
     $errors['passwordErr'] = "Password is required."; // add an error
}else{
     $password =($_POST['password']);
}

// etc etc

// check if there were any errors anywhere along the way
// and if not, proceed with login
if (!count($errors)) { // check there are no errors
    $sql="SELECT * FROM $tbl_name WHERE usrname='$username'";
    $result=mysql_query($sql);
    $count=mysql_num_rows($result);
    // etc etc
}else{
    // if there were errors do something else
    echo implode("<br />", $errors); // output the errors however you like
}

Answer 2

试一试

<?php

/* validate form first */
if (!empty($_POST['username']))
{ $username = $_POST['username'];
}
else{ echo "Username is required."; }

if (!empty($_POST['password']))
{ $password = $_POST['password'];
}
 else{ echo "password is required."; }


/* Do the queries second i.e */

SELECT * FROM Persons WHERE username='' AND password ='';



?>

Answer 3

嗨，您应该清楚地描述您的问题，我已经阅读了您的代码并进行了检查，当我没有填写密码时，它确实显示了Password is required. 一般验证方法如下：

if(empty($_POST['username'])){
    $usererror = '...';
    return false;
}else{
    $username = $_POST['username'];
}
if(empty($_POST['password'])){
     $passerror = '...';
     return false;
}else{
   $password = $_POST['password'];
}

Answer 4

处理错误验证的最佳方法是使用相同的变量，特别是如果您有许多输入表单数据

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

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if ($username == '') {
        $error_msg[]= 'Username is required';
    } else if ($password == '') {
        $error_msg[]= 'Password is required';
    }
}

if (!empty($error_msg)) {
    $ERROR_MSG = implode($error_msg);
    exit;
}

php：表单仍然在无效表单上提交

4 个答案: