Question

是否有人能够根据同一个表的另一个值来查找表的列值，从而为我提供MySQL关系查询？

例如，我有一个包含4列（id，name，email，password）的表。如何根据“email”列中的电子邮件查找某个用户的“id”列的值，并将结果（id）存储在变量中？

这是session（）控制器功能

function userIsLoggedIn()
{
    if (isset($_POST['action']) and $_POST['action'] == 'login')
    {
        if (!isset($_POST['email']) or $_POST['email'] == '' or
            !isset($_POST['password']) or $_POST['password'] == '')
        {
            $GLOBALS['loginError'] = 'Please fill in both fields';
            return FALSE;
        }

        $password = md5($_POST['password'] . 'chainfire db');

        if (databaseContainsAuthor($_POST['email'], $password))
        {
            session_start();
            $_SESSION['loggedIn'] = TRUE;
            $_SESSION['email'] = $_POST['email'];
            $_SESSION['password'] = $password;
            $_SESSION['authorid'] = $author;
            return TRUE;
        }
        else
        {
            session_start();
            unset($_SESSION['loggedIn']);
            unset($_SESSION['email']);
            unset($_SESSION['password']);
            unset($_SESSION['authorid']);
            $GLOBALS['loginError'] =
                    'The specified email address or password was incorrect.';
            return FALSE;
        }
    }

我试过

SELECT id FROM article WHERE email='$email'

有更有效的方法吗？我也不完全确定如何在$author会话变量中存储查询结果。

提前致谢。

Answer 1

您的查询是完美的（条件是您正在清理$email变量 - 请参阅Little Bobby Tables的故事。）

SELECT id FROM article WHERE email='$email'

要执行该查询并将变量$author设置为返回值，只需：

$author = 0;
if( ( $idFromEmail = @mysql_query( "SELECT id FROM article WHERE email='".mysql_real_escape_string( $email )."'" ) )
    && mysql_num_rows( $idFromEmail )==1
    && ( $r = mysql_fetch_assoc( $idFromEmail ) ) ){
    $author = $r['id'];
}

如果以上内容过于复杂而无法以长篇形式重写：

$author = 0;
if( !( $idFromEmail = @mysql_query( "SELECT id FROM article WHERE email='".mysql_real_escape_string( $email )."'" ) ){
  // SQL Query Failed
}elseif( mysql_num_rows( $idFromEmail )!=1 ){
  // More than, or less than, one row returned
}else{
  $r = mysql_fetch_assoc( $idFromEmail );
  $author = $r['id'];
}

根据用户电子邮件查询ID查找

1 个答案: