我最近搬到了另一台主机，并且遇到了问题。当我在网站表单中输入数据时，数据库不会更新。然而，当我的托管服务的运营商在他们这边做这件事时，该表格似乎对他们有用。为什么会这样？
这是 PHP 中的数据库调用
<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
$name = $_POST['name1'];
$surname = $_POST['surname1'];
$email = $_POST['email1'];
$email = filter_var($email, FILTER_SANITIZE_EMAIL); // Sanitizing E-mail.
if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
define('DB_NAME', 'name');
define('DB_USER', 'user');
define('DB_PASSWORD', 'password');
define('DB_HOST','localhost');
$conn = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
$name = mysqli_real_escape_string($conn, $name);
$surname = mysqli_real_escape_string($conn,$surname);
$email = mysqli_real_escape_string($conn,strtolower($email));
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO interested (Name, Surname, Email) VALUES ('$name', '$surname', '$email')";
if ($conn->query($sql) === TRUE) {
echo "Thank you!";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
} else {
echo "<span>* invalid email *" + $email +"</span>";
}
?>
最佳答案
我忘了提及，您还必须激活 mysql 级别的错误报告。例如。设置$mysqliDriver->report_mode，如下所示。
因此，在更改数据库凭据后，按原样测试此代码。如果有效，则问题出在您的代码中。如果没有，那么它就在其他地方。
<?php
// Db configs.
define('HOST', 'localhost');
define('PORT', 3306);
define('DATABASE', 'tests');
define('USERNAME', 'root');
define('PASSWORD', 'root');
/*
* Enable internal report functions. This enables the exception handling,
* e.g. mysqli will not throw PHP warnings anymore, but mysqli exceptions
* (mysqli_sql_exception).
*
* MYSQLI_REPORT_ERROR: Report errors from mysqli function calls.
* MYSQLI_REPORT_STRICT: Throw a mysqli_sql_exception for errors instead of warnings.
*
* @link http://php.net/manual/en/class.mysqli-driver.php
* @link http://php.net/manual/en/mysqli-driver.report-mode.php
* @link http://php.net/manual/en/mysqli.constants.php
*/
$mysqliDriver = new mysqli_driver();
$mysqliDriver->report_mode = (MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
// Create a new db connection.
$connection = new mysqli(HOST, USERNAME, PASSWORD, DATABASE, PORT);
// Signalize if a new record could be inserted, or not.
$recordInserted = FALSE;
// Operations upon form submission.
if (isset($_POST['submit'])) {
// Read the posted values.
$name = $_POST['name1'] ?? '';
$surname = $_POST['surname1'] ?? '';
$email = $_POST['email1'] ?? '';
// Validate the name.
if (empty($name)) {
$errors[] = 'Please provide the name.';
} /* Other validations here using elseif statements */
// Validate the surname.
if (empty($surname)) {
$errors[] = 'Please provide the surname.';
} /* Other validations here using elseif statements */
// Validate the email.
if (empty($email)) {
$errors[] = 'Please provide an email address.';
} elseif (
!filter_var($email, FILTER_SANITIZE_EMAIL) ||
!filter_var($email, FILTER_VALIDATE_EMAIL)) {
$errors[] = 'The email address is not in a valid format.';
} /* Other validations here using elseif statements */
// Insert a new record - if no errors yet.
if (!isset($errors)) {
$sql = 'INSERT INTO interested (
Name,
Surname,
Email
) VALUES (
?, ?, ?
)';
$statement = $connection->prepare($sql);
$statement->bind_param('sss', $name, $surname, $email);
$statement->execute();
// Signalize that the new record was successfully inserted.
$recordInserted = TRUE;
}
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" />
<meta name="viewport" content="width=device-width, initial-scale=1, user-scalable=yes" />
<meta charset="UTF-8" />
<!-- The above 3 meta tags must come first in the head -->
<title>Demo - Insert</title>
</head>
<body>
<div class="messages">
<?php
if (isset($errors)) {
echo implode('<br/>', $errors);
} elseif ($recordInserted) {
echo 'A new record was successfully inserted.';
}
?>
</div>
<br/>
<form action="" method="post">
<label for="name1">Name</label>
<input type="text" id="name1" name="name1" placeholder="Name" required>
<br/>
<label for="surname1">Surname</label>
<input type="text" id="surname1" name="surname1" placeholder="Surname" required>
<br/>
<label for="email1">Email</label>
<input type="email" id="email1" name="email1" placeholder="Email" required>
<button type="submit" id="submit" name="submit" value="submit">
Submit
</button>
</form>
</body>
</html>
关于php - SQL 调用 - 这是正确的吗？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51050369/