Question

这是我从表order和user_master

获取数据的SQL查询

String str = "select od.total, od.invoice_link, od.status, ";
str += "od.order_type, um.name as sender_name, umr.name as receiver_name, ";
str += "umr.twitter_handle as receiver_twitter_handle, um.twitter_handle as sender_twitter_handle ";
str += "from orders od ";
str += "join user_master um on um.id = od.user_id ";
str += "join user_master umr on umr.id = od.receiver_id ";
str += "where user_id = '"+user_id+"' ";

String json = dao.getResultJSON(str);
return json;

在数据库中有一个名为order_type的列，其中包含transfer，request和invoice等值。如果order_type列中的值为'transfer'且'Requested from'，则需要将“Payment to”返回到变量json中如果列中的值是'request'。请忘记我上面提到的'发票'。这两个就足以满足我的预期输出..

无论如何，提前谢谢

Answer 1

在不知道payment_to和requested_from应该来自何处的情况下，这是在黑暗中拍摄的：

select 
    od.total
  , od.invoice_link
  , od.status
  , od.order_type
  , um.name as sender_name
  , umr.name as receiver_name
  , um.twitter_handle as sender_twitter_handle 
  , umr.twitter_handle as receiver_twitter_handle
  , case when od.order_type = 'transfer' then od.payment_to else null end as payment_to 
  , case when od.order_type = 'request' then od.requested_from else null end as requested_from
  from orders od 
    inner join user_master um on um.id = od.user_id 
    inner join user_master umr on umr.id = od.receiver_id 
  where user_id = '"+user_id+"'

以您发布的格式：

String str = "select";
str += "    od.total";
str += "  , od.invoice_link";
str += "  , od.status";
str += "  , od.order_type";
str += "  , um.name as sender_name";
str += "  , umr.name as receiver_name";
str += "  , um.twitter_handle as sender_twitter_handle ";
str += "  , umr.twitter_handle as receiver_twitter_handle";
str += "  , case when od.order_type = 'transfer' then od.payment_to else null end as payment_to";
str += "  , case when od.order_type = 'request' then od.requested_from else null end as requested_from";
str += "  from orders od ";
str += "    inner join user_master um on um.id = od.user_id ";
str += "    inner join user_master umr on umr.id = od.receiver_id ";
str += "  where user_id = '"+user_id+"';";

String json = dao.getResultJSON(str);
return json;

Answer 2

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是从看似Java的SQL解析出来的。您应该了解SQL在表中处理;也就是说，它处理充满数据的矩形。每个结果集都返回一系列行。结果集中的每一行必须与其他行具有相同的列。

您要求提供两种不同的行，一行包含select od.total, od.invoice_link, od.status, od.order_type, um.name as sender_name, umr.name as receiver_name, umr.twitter_handle as receiver_twitter_handle, um.twitter_handle as sender_twitter_handle from orders od join user_master um on um.id = od.user_id join user_master umr on umr.id = od.receiver_id where user_id = SOMETHING列，另一行包含Payment to列。行的选择取决于行中的某些数据。这不会发生在单个查询中。

你可以做这样的事情。

Requested from

这将在结果集的每一行中生成所需的两列。当不相关时，列中的一列或两列将包含空格。

这些select ..., CASE WHEN od.order_type='transfer' THEN receiver_twitter_handle ELSE '' END `Payment to`, CASE WHEN od.order_type='request' THEN sender_twitter_handle ELSE '' END `Requested from`, ...语句的详细信息可能因表服务器的不同而有所不同，您还没有透露您正在使用哪一个，因此这里可能存在错误。但你可以解决它。

如何在SQL select语句

2 个答案: