將數據插入mysql數據庫（php，mysql）
[英]Inserting data in to mysql database (php, mysql)
問題描述
嗨，我正在使用 php 和 mysql 建立一個在線乳制品。 我嘗試注冊以便可以將用戶信息存儲到數據庫中，但我收到以下錯誤消息：“注意：未定義的變量：第 85 行 C:\wamp64\www\dairy\index.php 中的錯誤”我不知道如何解決這個問題。 請我需要幫助...
這是代碼：
<?php
if (array_key_exists("submit", $_POST)) {
$connection = mysqli_connect("localhost","root","","dairy_database");
if(mysqli_connect_error()) {
die ("There was an error connection to the database");
}
$error = "";
if (!$_POST['email']) {
$error .= "An email address is required <br>";
}
if (!$_POST['password']) {
$error .= "A password is required <br>";
}
if ($error != "") {
$error = "<p> There were error(s) in your form:</p>".$error;
}
else{
$query = "SELECT Id FROM users WHERE email = '".mysqli_real_escape_string($connection, $_POST['email'])."' LIMIT 1";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows($result) > 0) {
$error = "That email address is already taken.";
}
else{
$query = "INSERT INTO users ('email', 'password') VALUES ('".mysqli_real_escape_string($connection, $_POST['email'])."', '".mysqli_real_escape_string($connection, $_POST['password'])."')";
if (mysqli_query($connection, $query)) {
$error = "<p>Could not sign you up, Please try again later.</p>";
}
else{
$query = "UPDATE users SET password = '".md5(md5(mysqli_insert_id($connection)).$_POST['password'])."' WHERE Id = ".mysqli_insert_id($connection)." LIMIT 1";
mysqli_query($connection, $query);
echo "Sign up successful";
}
}
}
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<meta name="author" content="Bachir Amadou">
<link href="https://fonts.googleapis.com/css?family=Abel|Comfortaa" rel="stylesheet">
<link rel="stylesheet" href="../carousel/css/style.css" type="text/css">
<link rel="stylesheet" href="coverflow/css/style.css">
<title>Dairy</title>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"> </script>
<script type="text/javascript" src=""></script>
</head>
<body>
<div id="wrapper">
<div id="top">
<img class="log-img" src="./img/notebook.png" alt="notebook">
<h5>Online Dairy Management System</h5>
<h1>Web Base System</h1>
</div><!--END OF TOP-->
<div id="error"> <?php echo $error; ?></div>
<div id="signup">
<form id="form-signup" method="post">
<input type="email" class="user" name="email" placeholder="Email..."
required><br>
<input type="password" class="user" name="password" placeholder="Password" required><br>
<input type="submit" name="submit" class="sign-submit" value="Sign Up">
</form>
</div><!--END OF LOG-->
</div><!--END OF WRAPPER-->
</body>
</html>
2 個解決方案
解決方案1
0 2017-05-28 03:05:48
這是因為變量$error是在條件中聲明的。 您可以使用isset()函數來檢查變量是否設置。 一種替代解決方案是。
將此行更改為
<div id="error"> <?php echo $error; ?></div>
這個。
<?php if(isset($error)){ ?> <div id="error"> <?php echo $error; ?></div><?php } ?>
或者
在條件之前聲明變量，例如
<?php
$error = '';
if (array_key_exists("submit", $_POST)) {
....STATEMENTS.....
}
?>
<div id="error"> <?php echo $error; ?></div>
您在插入時的條件語句中也有一些錯誤。 我已經更新了下面的查詢。
PHP
<?php
$error = ""; // declare it here
if (array_key_exists("submit", $_POST)) {
$connection = mysqli_connect("localhost","root","","dairy_database");
if(mysqli_connect_error()) {
die ("There was an error connection to the database");
}
if (!$_POST['email']) {
$error .= "An email address is required <br>";
}
if (!$_POST['password']) {
$error .= "A password is required <br>";
}
if ($error != "") {
$error = "<p> There were error(s) in your form:</p>".$error;
}
else{
$query = "SELECT Id FROM users WHERE email = '".mysqli_real_escape_string($connection, $_POST['email'])."' LIMIT 1";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows($result) > 0) {
$error = "That email address is already taken.";
}
else{
$query = "INSERT INTO `users` (`email`, `password`) VALUES ('".mysqli_real_escape_string($connection, $_POST['email'])."', '".md5(md5(mysqli_real_escape_string($connection, $_POST['password'])))."');";
if (!mysqli_query($connection, $query)) {
$error = "<p>Could not sign you up, Please try again later.</p>";
}
else{
$query = "UPDATE `users` SET `password` = '".md5(md5(mysqli_insert_id($connection)).$_POST['password'])."' WHERE Id = ".mysqli_insert_id($connection)." LIMIT 1;";
mysqli_query($connection, $query);
echo "Sign up successful";
}
}
}
}
?>
HTML
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<meta name="author" content="Bachir Amadou">
<link href="https://fonts.googleapis.com/css?family=Abel|Comfortaa" rel="stylesheet">
<link rel="stylesheet" href="../carousel/css/style.css" type="text/css">
<link rel="stylesheet" href="coverflow/css/style.css">
<title>Dairy</title>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"> </script>
<script type="text/javascript" src=""></script>
</head>
<body>
<div id="wrapper">
<div id="top">
<img class="log-img" src="./img/notebook.png" alt="notebook">
<h5>Online Dairy Management System</h5>
<h1>Web Base System</h1>
</div><!--END OF TOP-->
<div id="error"> <?php echo $error; ?></div>
<div id="signup">
<form id="form-signup" method="post">
<input type="email" class="user" name="email" placeholder="Email..."
required><br>
<input type="password" class="user" name="password" placeholder="Password" required><br>
<input type="submit" name="submit" class="sign-submit" value="Sign Up">
</form>
</div><!--END OF LOG-->
</div><!--END OF WRAPPER-->
</body>
解決方案2
-1 2017-05-28 03:17:32
您需要在 if 條件之前聲明錯誤變量。 像這樣：
<?php
$error = ""; // declare it here
if (array_key_exists("submit", $_POST)) {
$connection = mysqli_connect("localhost","root","","dairy_database");
if(mysqli_connect_error()) {
die ("There was an error connection to the database");
}
if (!$_POST['email']) {
$error .= "An email address is required <br>";
}
if (!$_POST['password']) {
$error .= "A password is required <br>";
}
if ($error != "") {
$error = "<p> There were error(s) in your form:</p>".$error;
}
else{
$query = "SELECT Id FROM users WHERE email = '".mysqli_real_escape_string($connection, $_POST['email'])."' LIMIT 1";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows($result) > 0) {
$error = "That email address is already taken.";
}
else{
$query = "INSERT INTO users ('email', 'password') VALUES ('".mysqli_real_escape_string($connection, $_POST['email'])."', '".mysqli_real_escape_string($connection, $_POST['password'])."')";
if (mysqli_query($connection, $query)) {
$error = "<p>Could not sign you up, Please try again later.</p>";
}
else{
$query = "UPDATE users SET password = '".md5(md5(mysqli_insert_id($connection)).$_POST['password'])."' WHERE Id = ".mysqli_insert_id($connection)." LIMIT 1";
mysqli_query($connection, $query);
echo "Sign up successful";
}
}
}
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<meta name="author" content="Bachir Amadou">
<link href="https://fonts.googleapis.com/css?family=Abel|Comfortaa" rel="stylesheet">
<link rel="stylesheet" href="../carousel/css/style.css" type="text/css">
<link rel="stylesheet" href="coverflow/css/style.css">
<title>Dairy</title>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"> </script>
<script type="text/javascript" src=""></script>
</head>
<body>
<div id="wrapper">
<div id="top">
<img class="log-img" src="./img/notebook.png" alt="notebook">
<h5>Online Dairy Management System</h5>
<h1>Web Base System</h1>
</div><!--END OF TOP-->
<div id="error"> <?php echo $error; ?></div>
<div id="signup">
<form id="form-signup" method="post">
<input type="email" class="user" name="email" placeholder="Email..."
required><br>
<input type="password" class="user" name="password" placeholder="Password" required><br>
<input type="submit" name="submit" class="sign-submit" value="Sign Up">
</form>
</div><!--END OF LOG-->
</div><!--END OF WRAPPER-->
</body>
</html>
你也應該使用准備好的語句而不是 mysqlihttps://www.w3schools.com/php/php_mysql_prepared_statements.asp
數據未插入php中的mysql數據庫
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