-1
出於某種原因，我的驗證密碼不工作在我的PHP。不知道這是命令還是我錯過了一些東西。如果密碼不正確，我希望它不允許用戶登錄。登錄驗證密碼/確認密碼不起作用。 PHP/MySQL
下面的HTML和PHP。
<HTML>
<HEAD>
<TITLE> Programming </TITLE>
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css">
<LINK REL="stylesheet" TYPE="text/css" href="homework2.css">
</HEAD>
<BODY>
<!-- CSS for http://the7.dream-demo.com/ -->
<div id="container">
<div id="header">
<div class="menuitem"> <a href="home.html">Home</a> </div>
<div class="menuitem"><a href="products.html">Products</a></div>
<div class="menuitem"><a href="cases.html">Case Studies</a></div>
<div class="menuitem"><a href="pricing.html">Pricing</a></div>
<div class="menuitem"><a href="aboutus.html">About Us</a></div>
</div>
<div id="bodycontent">
<div id="banner">
<div id="bannerleft"> <h1> We make you better athletes. Find out how! </h1> </div>
<div id="signin">
<form class="well form-inline" action="login.php" method="post">
<input type="text" class="input-small" placeholder="Email" name="email" >
<input type="password" class="input-small" placeholder="Password" name="password">
<br><br>
<!--
If you do not want to use twitter bootstrap css then you should uncomment next 6 lines and uncomment the
above 2 lines that provide input boxes
\t <label for="email">Email:</label>
\t <input type="text" name="email" id="email">
\t <br>
\t <label for="password">Password:</label>
\t <input type="password" name="password" id="password">
\t <br>
-->
<input type="submit" name="submit" id="logmein" value="Log In">
</form>
</div>
</div>
<div id="featurestrip">
<div id="signup">
<form action="signup.php" method="post">
<label for="firstname">Firstname:</label>
<input type="text" name="signup-firstname" id="signup-firstname">
<br>
<label for="lastname">Lastname:</label>
<input type="text" name="signup-lastname" id="signup-lastname">
<br>
<label for="email">Email: </label>
<input type="text" name="signup-email" id="signup-email">
<br>
<label for="password">Password:</label>
<input type="password" name="signup-password" id="signup-password">
<br>
<label for="password">Reconfirm Password:</label>
<input type="password" name="signup-repassword" id="signup-repassword">
<br><br>
<input type="submit" name="signmeup" id="signmeup" value="Sign Me Up!">
</form>
</div>
<div id="featureright"> <p>Sign up and find out more on how we can help. Pricing starts at $19.95 a month. </p>
<p><h3>Premium service starts at $49.95.</h3></p>
</div>
</div>
<div id="corefeatures">
<img height="200px" src="http://www.hockeymanitoba.ca/wp-content/uploads/2013/02/ltad-model.jpg">
</div>
<div id="testimonials"> Testimonial
<img height="200px" src="http://www.neuroexplosion.com/storage/development%20model%20jpeg.jpg?__SQUARESPACE_CACHEVERSION=1305662626397">
<img height="200px" src="http://www.phecanada.ca/sites/default/files/physical_literacy/LTAD_FMS.jpg">
</div>
<!--
<div id="portfolio"> Portfolio</div>
<div id="skills"> Skills</div>
-->
</div>
<div id="footer">Copyright Notice. All Rights Reserved. 2014</div>
</div>
</BODY>
</HTML>PHP
<?php
$mysql_hostname = 'localhost';
$mysql_user = 'username';
$mysql_password = 'password';
$mysql_database = 'db_users2015';
$connect = mysql_connect($mysql_hostname, $mysql_user, $mysql_password)
or die ("Couldn't connect");
echo "<BR>Connection Successful";
//to put data into database
//select database
$db_selected= mysql_select_db($mysql_database, $connect)
or die ("Couldn't connect to the database");
$email= $_POST['email'];
$password= $_POST['password'];
$sql = "SELECT COUNT (*) FROM users WHERE email= '{$_POST['email']}' AND password= '{$_POST['password']}'";
$sql_result = mysql_query($sql);
if(["email"]==$email &&["password"]==$password)
echo ("Login Successful.");
else{
die("Wrong Password.");
}
$sql = "SELECT COUNT(*) FROM users WHERE email= '{$_POST['email']}'";
$sql_result = mysql_query($sql);
if (mysql_result($sql_result, 0)<1)
{
die("<BR>Email address not found");
}
else{
echo "Login Successful!";
}
?>
和昨天一樣的音符。您正在測試的密碼是否包含報價？還有什麼是[[「email」]和'[「password」]'？如果'$ sql_result'爲true，那麼你不需要那麼那麼電子郵件和密碼匹配。但不這樣做，用戶輸入與查詢分開。 – chris85
我看到你在那裏提到，我改變了引用來匹配，今天再次感謝你的幫助chris85。我試圖使用電子郵件和密碼來匹配數據庫中的電子郵件和密碼，以確保它是正確的登錄名。你是說我不需要在下面的部分添加這些變量？如果（$ sql_result [「email」] == $ email && $ sql_result [「password」] == $ password） echo（「Login Successful。」）; 其他{ die（「密碼錯誤」）; } – ruwadidr
是的，如果它在匹配的SQL中匹配，則不需要再次檢查......但是不要將用戶輸入直接傳遞給你的SQL;並且不要以明文形式存儲密碼至少爲md5。有更多的教程，線程和函數可供更好的方法使用。你應該更新到'mysqli'或'pdo'驅動程序。 – chris85