PHP MySQL列无效
在过去的四天里,我一直在努力让它工作。它只是一个简单的登录页面,并没有存储敏感信息,但我在PHP方面遇到了问题PHP MySQL列无效,php,mysql,mysql-error-1054,Php,Mysql,Mysql Error 1054,在过去的四天里,我一直在努力让它工作。它只是一个简单的登录页面,并没有存储敏感信息,但我在PHP方面遇到了问题 if ($_SERVER['REQUEST_METHOD'] == 'POST') { $uname = $_POST["login"]; $pword = $_POST["pass"]; $uname = htmlspecialchars($uname); $pword = htmlspecialchars($pword); $user_nam
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$uname = $_POST["login"];
$pword = $_POST["pass"];
$uname = htmlspecialchars($uname);
$pword = htmlspecialchars($pword);
$user_name = "bradf294_access";
$password = "********";
$database = "bradf294_clients";
$server = "localhost";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
print(mysql_errno());
print($db_found);
if(isset($db_found)){
print($db_found."Success");
$SQL = "SELECT * FROM basicinfo WHERE ref = $uname AND pass = $pword";
$result = mysql_query($SQL);
print("Query made");
print(mysql_errno());
if ($result) {
print("result:".$result);
}
else {
print("Incorrect Login Details");
}
if ($result > 0) {
print("found user");
$errorMessage= "logged on ";
session_start();
$_SESSION['login'] = "1";
header ("Location: progressuser.php");
}
else {
print("Invalid Logon");
}
} else {
print("Database not found. The Webmaster has been notified. Please try again later");
$subject = "Automated login error" ;
$message = "An error occured whilst trying to connect to the MySQL database, to login to the progress checker" ;
mail("a-bradfield@bradfieldandbentley.co.uk", $subject, $message);
}
从我用来调试的页面上的输出来看,似乎是那些行不起作用,它们给出了一个1054错误-“未知列“%s”在“%s”中”
即使我将$SQL
字符串复制并粘贴到phpMyAdmin中,并且它工作得很好
有没有什么明显的迹象表明我做错了?转到并输入详细信息参考:
TST001
并传递:dnatbtr121
以查看自己的输出。您的WHERE条件中的值是否应该像在普通MySQL语句中一样不被引号括起来?对此外,您还将得到一系列关于SQL注入的评论。您需要引用以下变量:
$SQL = "SELECT * FROM basicinfo WHERE ref = '$uname' AND pass = '$pword'";
但是
mysql\u*
函数已被弃用-您应该考虑改为使用PDO
或mysqli\u*
。这两种方法都可以让您更容易地编写安全代码,并为您解决报价问题。能否显示table basicinfo?您的PHP大括号不平衡。确保你的代码样本在语法上是正确的是很重要的!谢谢,你能推荐一些资源吗?php.net上有很多信息——PDO页面在,mysqli快速入门指南在thankyou,是的,我意识到它根本不安全!我还不打算在数据库上存储任何有价值的信息!
$SQL = "SELECT * FROM basicinfo WHERE ref = '$uname' AND pass = '$pword'";