在PHP中使用mysql_query顯示用戶名不起作用
[英]Using mysql_query in PHP to show the user name is not working
問題描述
我目前正在嘗試創建注冊表單，並且表單本身可以運行，人們可以在我的數據庫中創建用戶的表單，但是當他們注冊后，它將他們重定向到admin.php 。
他們用來創建帳戶的名稱不會顯示，只能按行用戶名顯示。 它應該顯示“ Welcome， user_name ，您現在已經登錄！”
我只是無法顯示名字，但其他一切正常！
警告：mysql_fetch_array（）期望參數1為資源，在第25行的C：\\ path \\ to \\ admin.php中給出布爾值
警告：mysql_fetch_array（）期望參數1為資源，在第36行的C：\\ path \\ to \\ login.php中給出布爾值
管理員：
<?php
require('db_config.php');
require_once('functions.php');
//if the cookie is still valid, recreate the session
if( $_COOKIE['logged_in'] == true ){
$_SESSION['logged_in'] = true;
$_SESSION['user_id'] = $_COOKIE['user_id'];
$_SESSION['is_admin'] = $_COOKIE['is_admin'];
}
if( $_SESSION['logged_in'] != true ){
//not logged in! send them back to the form]
header('location:login.php');
}
//extract the data for the logged in user, so we can use it on all page
$user_id = $_SESSION['name'];
$query_user = "SELECT * FROM users
WHERE name = $user_id
LIMIT 1";
$result_user = mysql_query($query_user);
$row_user = mysql_fetch_array($result_user);
//this going to be a handy variable to have throughout all pages
$user_id = $row_user['user_id'];
?>
<!doctype HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta charset="utf-8">
<link rel="stylesheet" type="text/css" href="css/reset.css" />
<link rel="stylesheet" type="text/css" href="css/format.css" />
<title>Schell Shock Design's Portfolio</title>
</head>
<body>
<div id="login">
<?php
include('login.php');
?>
</div>
<div id="utilities">
<?php include('utilities.php'); ?>
</div>
<div id="container">
<header>
<?php include('header.php'); ?>
</header>
<div id="slider">
<?php include('slider.php'); ?>
</div>
<div id="content">
<?php include('content.php'); ?>
</div>
<div id="bottomcontent">
<?php include('bottomcontent.php'); ?>
</div>
<div id="footer">
<?php include('footer.php'); ?>
</div>
</body>
</html>
登錄：
<?php
//show an error if there is a problem with the login
if($error == true){ ?>
<div class="error">
Sorry, Your username and password are incorrect. Try again.
</div>
<?php } //end if error ?>
<?php //show the form only if NOT logged in
if( !$_SESSION['logged_in'] ){
?>
<div class="form1">
<form action="?action=" method="post">
<label for="username">Username:</label>
<input type="text" name="username" id="username" />
<label for="password">Password</label>
<input type="password" name="password" id="password" />
<input type="submit" value="Log in" />
<input type="hidden" name="did_login" value="1" />
</form>
<?php } //end if not logged in
else{
//get info of logged in person
$user_id = $_SESSION['user_id'];
$query_user = "SELECT name
FROM users
WHERE user_id = $user_id";
$result_user = mysql_query( $query_user );
$row_user = mysql_fetch_array( $result_user );
?>
<div id="loggedin">
<a href="?action=logout">Log Out</a>
<?php //show a welcome message if they logged in successfully
echo 'Welcome '.$row_user['name'].', You are now logged in!';
?>
<?php } ?>
</div>
注冊
<?php
//register parse. all this logic MUST go before the doctype or any other text output.
require('db_config.php');
require_once('functions.php');
//if they submitted the form, parse it
if( $_POST['did_register'] == 1 ){
//extract amd sanitize all fields
$username = clean_input($_POST['username']);
$email = clean_input($_POST['email']);
$password = clean_input($_POST['password']);
$repassword = clean_input($_POST['repassword']);
$policy = clean_input($_POST['policy']);
//encrypted version of the password, for storing in the database
$sha_password = sha1($password);
//begin validation
$valid = true;
//did they forget to check the box?
if( $policy != 1 ){
$valid = false;
$msg = 'You must agree to the TOS and PP before signing up. <br />';
}
//repeated password does not match
if( $password != $repassword ){
$valid = false;
$msg .= 'The passwords provided do not match. <br />';
}
//make sure the username and password are at least 5 characters long, than check the database
if( strlen($username) >= 5 AND strlen($password) >= 5 ){
//check to see if username is already taken
$query_username = "SELECT name
FROM users
WHERE name = '$username'
LIMIT 1";
$result_username = mysql_query($query_username);
//if one result is found, username is taken.
if( mysql_num_rows($result_username) == 1 ){
$valid= false;
$msg .= 'That username is already taken. Try another. <br />';
}
}else{
$valid = false;
$msg .= 'Username and Password must be at least 5 characters long. <br />';
}
//check for valid email, than check for match in database
if( check_email_address($email) == true ){
//look for match in database
$query_email = "SELECT email
FROM users
WHERE email = '$email'
LIMIT 1";
$result_email = mysql_query($query_email);
//if 1 result is found, email is taken.
if( mysql_num_rows($result_email) == 1 ){
$valid = false;
$msg .= 'Looks like an account with that email already exists. Do you want to login? <br />';
}
}else{
//invalid email
$valid = false;
$msg .= 'Please provide a valid email address. <br />';
}
//if the data passed ALL tests, add the user to the database
if( $valid == true ){
$query_insert = "INSERT INTO users
(name, password, email, join_date, is_admin)
VALUES
('$username', '$sha_password', '$email', now(), 0)";
$result_insert = mysql_query($query_insert);
//check to see if it worked
if( mysql_affected_rows() == 1 ){
//SUCCESS! Log the user in and send them to their profile.
$_SESSION['logged_in'] = true;
setcookie( 'logged_in', 'true', time() + 60*60*24*7 );
header( 'location:index.php' );
}else{
$msg .= 'There was a problem adding the user to the Database';
}
}
} //end if submitted form
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Sign up for an account</title>
</head>
<body>
<?php
if( isset($msg) ){
echo $msg;
}
?>
<form action="registration.php" method="post">
<label for="username">Choose a Username:</label>
<input type="text" name="username" id="username" />
<span class="hint">Minimum of five characters</span>
<label for="email">Your Email Address:</label>
<input type="text" name="email" id="email" />
<label for="password">Choose a Password:</label>
<input type="password" name="password" id="password" />
<span class="hint">Minimum of 5 characters</span>
<label for="repassword">Repeat Password:</label>
<input type="password" name="repassword" id="repassword" />
<input type="checkbox" name="policy" id="policy" value="1" />
<label for="policy">Yes, I have read the Terms of Service and Privacy Policy.</label>
<input type="submit" value="Sign up" />
<input type="hidden" name="did_register" value="1" />
</form>
</body>
</html>
我需要解決什么？
2 個解決方案
解決方案1
1 2012-07-07 21:07:30
您應該檢查錯誤是什么：
if (!$result_user) { die('MySQL Error: '.mysql_error()); }在每個頁面的頂部調用
session_start()。並確保正確返回會話的值：
print_r($_SESSION);
解決方案2
0 2012-07-07 21:06:07
在admin.php ，此查詢失敗：
$query_user = "SELECT * FROM users WHERE name = $user_id LIMIT 1";
也許$user_id為空，或者需要用引號（ '$user_id' ）。
無論如何，您都應檢查查詢結果以確保查詢成功：
$user_id = $_SESSION['name'];
$query_user = "SELECT * FROM users
WHERE name = $user_id
LIMIT 1";
$result_user = mysql_query($query_user);
if (!$result_user) {
die('Query failed: ' . mysql_error());
}
mysql_query（）僅在成功時返回資源結果。 失敗時，它返回(bool)FALSE ，它不能傳遞給任何mysql_fetch_*函數。
login.php中的錯誤也是如此。
您似乎沒有顯示登錄時運行的代碼，我想您沒有為會話分配正確的變量。
使用mysql_query的innodb mysql / php中的事務回滾不起作用
[英]Transaction rollback in innodb mysql/php with mysql_query not working
聲明:本站的技術帖子網頁，遵循CC BY-SA 4.0協議，如果您需要轉載，請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.