將用戶信息插入mysql數據庫
[英]Insert user info into mysql database
問題描述
我為移動設備創建了一個社交網絡，並嘗試將一些用戶信息和文件“ settings.php”一起插入我的數據庫。 但是，當我運行“ settings.php”時，chrome給我顯示錯誤（我在顯示錯誤的文件中注釋。有人提示我應該更改什么嗎？
我的數據庫：
CREATE TABLE `users` (
`id` INT NOT NULL AUTO_INCREMENT ,
`username` VARCHAR NOT NULL ,
`password` VARCHAR NOT NULL ,
`picture` VARCHAR NOT NULL ,
`age` INT NOT NULL ,
`residence` VARCHAR NOT NULL ,
`status` VARCHAR NOT NULL ,
`sex` VARCHAR NOT NULL ,
的login.php
<?php
if (!empty($_POST)) {
if (
empty($_POST['f']['username']) ||
empty($_POST['f']['password'])
) {
$message['error'] = 'Es wurden nicht alle Felder ausgefüllt.';
} else {
$mysqli = @new mysqli('localhost', 'root', 'pw', 'database');
if ($mysqli->connect_error) {
$message['error'] = 'Datenbankverbindung fehlgeschlagen: ' . $mysqli->connect_error;
}
$query = sprintf(
"SELECT username, password FROM users WHERE username = '%s'",
$mysqli->real_escape_string($_POST['f']['username'])
);
$result = $mysqli->query($query);
if ($row = $result->fetch_array(MYSQLI_ASSOC)) {
if (crypt($_POST['f']['password'], $row['password']) == $row['password']) {
session_start();
$_SESSION = array(
'login' => true,
'user' => array(
'username' => $row['username']
)
);
$message['success'] = '';
header('Location: http://' . $_SERVER['HTTP_HOST'] . '/anyask/main.php');
}
}
$mysqli->close();
}
}
?>
<html>
<head>
<meta charset="UTF-8" />
<title>
HTML Document Structure
</title>
<link rel="stylesheet" type="text/css" href="style1.css" />
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.0/jquery.mobile-1.3.0.min.css" />
<script src="http://code.jquery.com/jquery-1.8.2.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.0/jquery.mobile-1.3.0.min.js"></script>
<meta name="apple-mobile-web-app-capable" content="yes" />
<meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">
</head>
<body>
<div id="wrapper">
<form name="login-form" class="login-form" action="./login.php" method="post">
<div class="header">
</div>
<div class="content">
<label for="username"></label>
<input name="f[username]" type="text" class="input username" placeholder="Username" id="username"
<?php
echo isset($_POST['f']['username']) ? ' value="' . htmlspecialchars($_POST['f']['username']) . '"' : '' ?>/>
<label for="password"></label>
<input name="f[password]" type="password" class="input password" placeholder="Password" id="password" />
</div>
<div class="footer">
<input type="submit" name="submit" value="Login" class="button" data-theme="b"/>
<a href="./register.php">Register</a>
</div>
</form>
</div>
<div class="gradient"></div>
</body>
</html>
auth.php
<?php
session_start();
session_regenerate_id();
if (empty($_SESSION['login'])) {
header('Location: http://' . $_SERVER['HTTP_HOST'] . '/login.php');
exit;
} else {
$login_status = '
<div style="border: 1px solid black">
Sie sind als <strong>' . htmlspecialchars($_SESSION['user']['username']) . '</strong> angemeldet.<br />
<a href="./logout.php">Sitzung beenden</a>
</div>
';
}
?>
的settings.php
<?php require_once './auth.php'; ?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<title>main</title>
<link rel="stylesheet" type="text/css" href="mainstyle.css" />
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.0/jquery.mobile-1.3.0.min.css" />
<script src="http://code.jquery.com/jquery-1.8.2.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.0/jquery.mobile-1.3.0.min.js"></script>
<meta name="apple-mobile-web-app-capable" content="yes" />
<meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1">
</head>
<body>
<div data-role="header" data-theme="b" data-position="fixed" data-tap-toggle="false" >
<h1 class="ui-title" role="heading" aria-level="1">anyask</h1>
</div>
<div data-role="main" class="ui-content">
<div id="wrapper">
<form name="login-form" class="login-form" action="./settings.php" method="post">
<div class="header">
</div>
<div class="content">
<label for="age"></label>
<input name="age" type="number" class="input age" placeholder="age" id="age"/>
<label for="residence"></label>
<input name="residence" type="text" class="input residence" placeholder="residence" id="residence" />
<fieldset data-role="controlgroup" data-type="horizontal">
<legend>sex:</legend>
<input type="radio" name="radio-choice-h-2" id="radio-choice-h-2a" value="man" checked="checked">
<label for="radio-choice-h-2a">man</label>
<input type="radio" name="radio-choice-h-2" id="radio-choice-h-2b" value="woman">
<label for="radio-choice-h-2b">woman</label>
</fieldset>
profile picture:
<input name="attachment" type="file" id="attachment" /><br>
</div>
<div class="footer">
<input type="submit" name="submit" value="save" class="button" data-theme="b"/>
</div>
</form>
</div>
<div class="gradient"></div>
</div>
</body>
</html>
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// upload picture
// the following 3 lines are showed as error
$age=$_POST['age'];
$residence=$_POST['residence'];
$sex=$_POST['radio-choice-h-2'];
// Insert data into mysql
$sql="INSERT INTO $tbl_name(age, residence, radio-choice-h-2)VALUES('$age', '$residence', '$sex')";
$result=mysql_query($sql);
// if successfully insert data into database, displays message "Successful".
if($result){
header('Location: http://' . $_SERVER['HTTP_HOST'] . '/anyask/main.php');
}
else {
echo "ERROR";
}
// close connection
mysql_close();
?>
1 個解決方案
解決方案1
1 已采納 2015-01-18 19:46:57
加載頁面后，$ _ POST變量中不存在變量age，駐留和radio-choice-h-2，因為這些變量僅在提交表單后存在。 （ http://php.net/manual/en/reserved.variables.post.php ）
您應該在運行數據庫查詢之前檢查這些變量是否存在。 像這樣：
if (isset($_POST['age'])) {
// DO SOMETHING
}
通常，您應該檢查變量是否存在並且包含正確的值。 特別是在使用表格時。
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