我正在讓用戶輸入他們的電子郵件和密碼,並且我希望從表中回顯他們的用戶名。這是我可以證明我已經從表中檢索到一個值。使用MySQL和PHP返回單個值
我知道我的查詢在phpmyadmin上工作,並且我的php中也檢索到了電子郵件和密碼。我與我當前的代碼得到的錯誤是HTTP錯誤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");
如果你有適當的訪問權限,您可以通過啓用錯誤報告對PHP來幫助自己。將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
也許可以幫助https://www.lifewire.com/500-internal-server-error-explained -2622938 – b2ok