이름과 라디오를 확인해야하므로 오류 코드가 작성되었습니다. 그렇지 않으면 확인 페이지로 이동하여 필드 옆에 오류 메시지를 보낼 수 없습니다. 코드를 놓친 경우 도움주세요.PHP 유효성 검사 오류
<?php
$nameErr = $charityErr = "";
$name = $charity = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$nameErr = "Name is missing";
}
else {
$name = $_POST["name"];
}
if (!isset($_POST["charity"])) {
$charityErr = "You must select 1 option";
}
else {
$charity = $_POST["charity"];
}
}
?>
<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link type="text/css" href="testStyle.css" rel="stylesheet"/>
<title>Survey</title>
</head>
<body>
<div><!--start form-->
<form method="post" action="submit.php">
<input type="radio" name="charity" <?php if (isset($charity) && $charity == "1") echo "checked"; ?> value="1">1
<input type="radio" name="charity" <?php if (isset($charity) && $charity == "2") echo "checked"; ?> value="2">2
<input type="radio" name="charity" <?php if (isset($charity) && $charity == "3") echo "checked"; ?> value="3">3
<span class="error"><?php echo $charityErr;?></span>
<input type="text" name="name" placeholder="ENTER YOUR COMPANY NAME">
<span class="error"><?php echo $nameErr;?></span>
<input type="submit" name="submit" value="Submit"/>
</form>
</div><!--end form-->
</body>
</html>
내 submit.php는 말한다 : 이것은 내 유효성 검사가 작동하지 않습니다 제외하고 페이지를 submit.php 괜찮 리디렉션 및 빈 데이터를 전송
/* Subject and Email Variables */
$emailSubject = 'Survey!';
$webMaster = '[email protected]';
/* Gathering Data Variables */
$name = $_POST ['name'];
$charity = $_POST ['charity'];
//create a new variable called $body line break, say email and variable, don't leave any space next to EOD - new variable $body 3 arrows less than greater than, beyond EOD no spaces
$body = <<<EOD
<br><hr><br>
Company Name: $name <br>
Charity: $charity <br>
EOD;
//be able to tell who it's from
$headers = "From: $email\r\n";
$headers .= "Content-type: text/html\r\n";
$success = mail($webMaster, $emailSubject, $body, $headers);
/* Results rendered as HTML */
$theResults = <<<EOD
<html>
blah blah
</html>
. 당신은 여기에 구문 오류가
<form method="post" action="submit.php">
<input type="radio" name="charity"
<?php if (isset($charity) && $charity == "1") echo "checked"; ?>
value="1">1
<input type="radio" name="charity"
<?php if (isset($charity) && $charity == "2") echo "checked"; ?>
value="2">2
<input type="radio" name="charity"
<?php if (isset($charity) && $charity == "3") echo "checked"; ?>
value="3">3
<span class="error"><?php echo $charityErr;?></span>
<input type="text" name="name" placeholder="ENTER YOUR COMPANY NAME">
<span class="error"><?php echo $nameErr;?></span>
<input type="submit" name="submit" value="Submit"/>
</form>
확신 라디오의 약간 다른 나와. ' php print $ charity;는 무엇을 하는가? ?>'양식을 게시 한 후에 출력 하시겠습니까? –
'echo '
';'submit.php를 출력 할 수 있습니까? – skobaljic어떻게 라디오 결과를 이메일로 보내나요? – Violin