Question

我是PHP的新手。我有这个非常基本的表单，我正在尝试添加非常基本的验证。我不知道我做错了什么，但是一旦我提交了我的表格，每个字段填满，它就不会去另一页。它将名称显示为未填充字段。

<?php
error_reporting(0);
if($_POST['submit'])
{
$error=array();
$n=$_POST['name'];
$p=$_POST['password'];
    if($n == '');
    {
        $error[0]= "Please fill a name";
    }
    if($p == '')
    {
        $error[1]= "Password is required";
    }

}
?>
<html>
<form method="post" action="<?php  if($_POST['submit']) { if(count($error)<1) { echo "hello.php"; } else { echo ''; } } ?>">
<div>
<span>Username:</span> <input type="text" name="name" id="user" /> <span style="color:red;"> <?php echo $error[0];?> </span>
</div>
<div>
<span>Password:</span><input type="password" name="password" id="pass" /><span style="color:red;"> <?php echo $error[1];?> </span>
</div>
<div>
<input type="submit" name="submit" value="submit" />
<input type="reset" name="reset" value="reset" />
</div>
</form>
</html>

现在即使当我提交表格时，每个字段都填满了我得到名字字段没有填写！这可能是一些愚蠢的事情，但我无法弄明白。

Answer 1

if($n == '');从此行中移除; 应该是

if($n == '')

这是您在评论中提出的问题。 //而不是编码动作。在表单中，如果$ error消息为空，则应检查它们。您可以将用户定向到hello.php。感谢

<?php error_reporting(0); if($_POST['submit']) { $error=array(); $n=$_POST['name']; $p=$_POST['password']; if(empty($n)) { $error[0]= "Please fill a name"; } if(empty($p)) { $error[1]= "Password is required"; } if(empty($error)) { header('Location:hello.php'); } } ?> <form method="post" action=""> <div> <span>Username:</span> <input type="text" name="name" id="user" /> <span style="color:red;"> <?php echo $error[0];?> </span> </div> <div> <span>Password:</span><input type="password" name="password" id="pass" /><span style="color:red;"> <?php echo $error[1];?> </span> </div> <div> <input type="submit" name="submit" value="submit" /> <input type="reset" name="reset" value="reset" /> </div> </form> </html>

Answer 2

在你的php中使用以下内容：

<?php
function errors($error) {
    echo '<ul> class="error"';
    foreach($error as $fail) {
        echo '<li>'.$fail.'</li>';
    }
    echo '</ul>';
}

// Sanitize user input
function sanitize($value) {
    // You need to pass your db connection to mysqli functions.
    // You can do this by setting it as a paramater to the function or include a file with the connection in the function (not really recommended)
    return mysqli_real_escape_string($connect, trim(strip_tags($value)));
}

//redirect a user
function to($url) {
   if(headers_sent()) {
       echo '<script>window.location("'.$url.'");</script>';
   } else {
       header('Location: '.$url);
       exit();
   }
}

// The code is only tiggered when the page is posted
if($_POST) {
    $error = array(); // store all errors
    $name = sanitize($_POST['name']);
    $pass = sanitize($_POST['password']);

    if(!empty($name) && !empty($pass)) {
        // continue with more validation
    } else {
        $error[] = 'You didn\'t enter a username and or password';
    }

    if(!empty($error)) {
        echo errors($error);
    } else {
        // if there are no errors (forexample log the user in)
        to('hello.php');
    }
}
?>

作为你的html，这就足够了：

<style>
.error ul {
	list-style-type: none;
	margin: 0;
	padding: 0;
}

.error ul li {
	color: red;
}
</style>

<html>
	<form action="" method="post">
		<div>
			<span>Username:</span> <input type="text" name="name" id="user">
		</div>
		<div>
			<span>Password:</span> <input type="password" name="password" id="pass">
		</div>
		<div>
			<input type="submit" value="Submit">
			<input type="reset" value="Reset">
		</div>
	</form>
</html>

一些有用的链接：

Answer 3

您可以使用php函数empty（），如

if(empty($n)){
$error[0]= "Please fill a name";
}

表单验证无法正常工作：名称仍未填充

3 个答案: