小编h3g*_*0r_的帖子

我需要一个简单的表格列.

例如,一个表"项目",带有列id,name和year.

如果我做:

$q = Doctrine_Query::create()
        ->select('a.pro_id')
        ->from('fndr_proyecto a')
        ->where('a.pro_id =?',1);
    $pro = $q->execute();
    json_encode($pro->toArray());

答案就是所有专栏

{"id":1,"name":"Project name","year":2013}

但我只需要一列.我预计:

{"id":1}

它与DQL一起使用,因为本机SQL工作正常.

ORM是使用Visual Paradigm自动构建的.

我需要得到所有状态(1,2,3,4)没有重复.

1 xxx
2 yyy
3 zzz
4 jjj

我的第一个也是唯一的想法就是让所有的州和他们分组:

select state1 as state,desc1 as desc from table where id=X
Union
select state2 as state,desc2 as desc from table where id=X

这样就可以获得6行示例.因此,为了丢弃副本,我尝试使用别名:

select state,desc from 
(
select state1 as state,desc1 as desc from table where id=X
    Union
    select state2 as state,desc2 as desc from table where id=X
)
group by state;

但我得到的错误不是GROUP BY表达式.

我看到了类似的问题,但我无法解决问题.