1

我在 postgres 中使用交叉表函数。基本 SQL 是:

select distinct  
       o_location,
       co_name,
       o_date,
       o_ndate,
       o_day,
       o_hour,
       o_type
  from outputs_txt
  left join courses on o_course = co_foreign
  left join locations on o_location = l_code
 where o_date = '2011-10-10' 
 order by o_hour

我的交叉表查询是

SELECT *
  FROM crosstab(
'SELECT DISTINCT
        COALESCE(o_location, '''')
       ,o_hour AS hour
       ,c.co_name
   FROM outputs_txt AS d
   LEFT JOIN courses AS c 
     on o_course = c.co_foreign
   LEFT JOIN locations as a 
     on o_location = a.l_code
  WHERE d.o_date = ''2011-10-10'''
)
AS ct(
 o_location varchar
,hour_0  varchar
,hour_1  varchar
,hour_2  varchar
,hour_3  varchar
,hour_4  varchar
,hour_5  varchar
,hour_6  varchar
,hour_7  varchar
,hour_8  varchar
,hour_9  varchar
,hour_10 varchar
,hour_11 varchar
,hour_12 varchar
,hour_13 varchar
,hour_14 varchar
,hour_15 varchar
,hour_16 varchar
,hour_17 varchar)

问题是结果都向左移动了。
例如,如果某个位置的课程应显示在hour_8其上,则会显示在 中hour_0
这适用于所有地点的所有课程。它们都左对齐。

我哪里错了?

4

1 回答 1

1

由于没有样本数据,我只能猜测。这让我印象深刻:您的交叉表函数
中没有ORDER BY。我加了一个:

SELECT *
  FROM crosstab(
'SELECT DISTINCT
        COALESCE(o_location, '''')
       ,o_hour AS hour
       ,c.co_name
   FROM outputs_txt AS d
   LEFT JOIN courses AS c    ON o_course = c.co_foreign
   LEFT JOIN locations as a  ON o_location = a.l_code
  WHERE d.o_date = ''2011-10-10''
  ORDER BY 1,2'
)
AS ct(
 o_location varchar
,hour_0  varchar
,hour_1  varchar
,hour_2  varchar
,hour_3  varchar
,hour_4  varchar
,hour_5  varchar
,hour_6  varchar
,hour_7  varchar
,hour_8  varchar
,hour_9  varchar
,hour_10 varchar
,hour_11 varchar
,hour_12 varchar
,hour_13 varchar
,hour_14 varchar
,hour_15 varchar
,hour_16 varchar
,hour_17 varchar)

如果你想hour_0先,你必须订购o_hour。我引用了交叉表函数的手册

在实践中,SQL 查询应始终指定 ORDER BY 1,2 以确保输入行正确排序,也就是说,将具有相同 row_name 的值放在一起并在行内正确排序。

于 2011-10-11T08:10:39.710 回答