Php 查询基于用户电子邮件的id查找

有人能给我一个MySQL关系查询，根据同一个表的另一个值查找表的列值吗

例如，我有一个包含4列（id、名称、电子邮件、密码）的表。如何根据某个用户的电子邮件在“电子邮件”列中查找“id”列的值，并将结果（id）存储在变量中

下面是session（）控制器函数

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;
        }
    }

我试过了

SELECT id FROM article WHERE email='$email' 

有没有更有效的方法？我也不完全确定如何将查询结果存储在

$author

会话变量中

提前谢谢

您的查询是完美的（附带条件是您正在清理

$email

变量-请参阅的故事）

要执行该查询并将变量

$author

设置为返回值，只需执行以下操作：

$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'];
}

如果上述内容过于复杂而无法遵循，则以长格式重写：

$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'];
}

---

我建议您将每段代码都写在单独的一行中，以使代码更具可读性，即，

$idFromMail=mysql\u query（…）；如果（mysql_num_rows（…）==1）{$r=mysql_fetch…；$author=$r['id']；}

同意。通常我会，我只是想在这里尝试一些新的东西。

$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'];
}