用zend框架编写mysql查询

我有一个很好的查询

SELECT t.username
FROM users t LEFT JOIN friends y ON t.id=y.user_id2 and y.user_id1=2
WHERE LOWER(t.username) LIKE 'ha%'
ORDER BY 
    CASE WHEN y.user_id2 IS NULL THEN 1 
    ELSE 0 
    END     
    ,t.username;

我正试图用zend框架编写它，这就是我想到的

        $users = new Users;
        $select = $users->select();
        $select->setIntegrityCheck(false);
        $select->from(array('t1' => 'users'), array('username'));
        $select->joinLeft(array('t2' => 'friends'), 't1.id=t2.user_id2 and t2.user_id1 =2');
        $select->where("LOWER(t1.username) like '$input%'");
        $select->order("t1.username, CASE WHEN t2.user_id2 IS NULL THEN 1 ELSE 0 END ");
        $listofusernames = $users->fetchAll($select);

然而，它似乎不工作，我得到这个错误

Fatal error: Uncaught exception 'Zend_Db_Statement_Mysqli_Exception' with message 'Mysqli prepare error: Unknown column 't1.username, CASE WHEN t2.user_id2 IS NULL THEN 1 ELSE 0 END ' in 'order clause'' in /opt/lampp/htdocs/vote_old/library/Zend/Db/Statement/Mysqli.php:77 Stack trace: #0

显然，这与order by条款中嵌入的案例有关

你知道如何修复那个代码吗

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谢谢

尝试将列放入数组

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致命错误：未捕获异常“Zend_Db_Statement_Mysqli_exception”并显示消息“Mysqli prepare error:Unknown column”如果t2.user_id2为空，则在“order子句”中“您需要使用Zend_Db_Expr，就像我不确定这是什么意思一样。”。你能说得更具体些吗？

$select->order(array('t1.username',
              new Zend_Db_Expr ('CASE WHEN t2.user_id2 IS NULL THEN 1 ELSE 0 END')));