我正在让用户输入他们的电子邮件和密码,并且我希望从表中回显他们的用户名。这是我可以证明我已经从表中检索到一个值。使用MySQL和PHP返回单个值
我知道我的查询在phpmyadmin上工作,并且我的php中也检索到了电子邮件和密码。我与我当前的代码得到的错误是HTTP错误500:
mysql_connect($database,$username,$password);
@mysql_select_db($username) or die ("Unable to select database");
$email = mysql_escape_string ($_POST['email']);
$password = mysql_escape_string ($_POST['password']);
$email = stripslashes($email);
$password = stripslashes($password);
$email = mysql_escape_string($email);
$password = mysql_escape_string($password);
$SQL = "SELECT Name FROM users WHERE BINARY Email = '$email' and BINARY password = '$password'";
$result = mysql_query($SQL) or die("Unable to Run Query");
$value = mysql_fetch_object($result) or die("Unable to Fetch Object");
echo "<h2>$value</h2>" or die("Unable to Echo");
如果你有适当的访问权限,您可以通过启用错误报告对PHP来帮助自己。将display_errors = on shouold添加到您的PHP.ini文件中。 http://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display – Stese
也许可以帮助https://www.lifewire.com/500-internal-server-error-explained -2622938 – b2ok