저는 PHP를 처음 사용하면서도 계속 저의 머리를 쓰려고합니다. 이 양식은 데이터가 데이터베이스로 전송되었다고 말하지만 데이터베이스가 비어있는 것처럼 보이면 오류가 나타나지 않습니까? 내 코드에 문제가 있습니까?보내지는 말은 있지만 데이터가 없습니다. 데이터베이스에 나타납니다
참고 :이 양식은 SQL Injection으로 보호되지 않습니다.
HTML
<?php
session_start();
?>
<!DOCTYPE HTML>
<html>
<head>
<title>Page Form</title>
<link rel="stylesheet" href="style.css" />
</head>
<body>
<div class="container">
<div class="main">
<h2>PHP Page 3 Form</h2><hr/>
<span id="error">
</span>
<form action="page4_insertdata.php" method="post">
<label>Company Name :<span>*</span></label><br />
<input name="company_name" type="text" placeholder="Joes Cleaner" required>
<br />
<label>Ref :<span>*</span></label><br />
<input name="ref" type="text" placeholder="H123" required>
<br />
<label>Website :<span>*</span></label><br />
<input name="website" type="text" placeholder="www.google.com" required>
<br />
<label>Email :<span>*</span></label><br />
<input name="email" type="email" placeholder="[email protected]" required>
<br />
<label>Telephone :<span>*</span></label><br />
<input name="tel" type="text" placeholder="07123456789" required>
<br />
<label>Message :<span>*</span></label><br />
<input name="message" id="message" type="text" size="500" required>
<br />
<input type="reset" value="Reset" />
<input name="submit" type="submit" value="Submit" />
</form>
</div>
</div>
</body>
</html>
PHP
<?php
session_start();
?>
<!DOCTYPE HTML>
<html>
<head>
<title>PHP Multi Page Form</title>
<link rel="stylesheet" href="style.css" />
</head>
<body>
<div class="container">
<div class="main">
<h2>PHP Multi Page Form</h2><hr/>
<?php
$servername = "localhost";
$db_database = 'form';
$username = "root";
$password = "";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "DB Connected successfully. ";
$company_name = $_POST['company_name'];
$ref = $_POST['ref'];
$website = $_POST['website'];
$email = $_POST['email'];
$tel = $_POST['tel'];
$message = $_POST['message'];
$sql = "INSERT INTO detail (company_name,ref,website,email,tel,message)
VALUES ('$company_name','$ref','$website','$email','$tel','$message')";
if($sql){
echo " Database Sent.";
}
else {
echo "ERROR to insert into database";
};
?>
</div>
</div>
</body>
</html>
:
당신은 할 필요가있다. –
** 위험 ** : [방어벽] (http://stackoverflow.com/questions/)이 필요한 [SQL 주입 공격] (http://bobby-tables.com/) ** **에 취약합니다. ** 60174/best-way-to-prevent-sql-injection-in-php)을 사용하십시오. – Quentin
나는 먼저 그것을 작동 시키려고 노력하는 Im을 안다. – jl5660