mysql - 使用 zend 框架编写 mysql 查询

Question

我有这个查询工作正常

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;

我正在尝试用 zend 框架编写它，这就是我想出的

        $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);

但是它似乎不起作用，我得到了这个错误

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

显然它与嵌入在 order by 子句中的大小写有关。

您知道如何修复该代码吗？

谢谢

Answer 1

尝试将列放入数组 like

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