Question

我正在撰写一份报告，显示自去年以来每个月的代码总量。

目前，如果我只计算过去一年中的所有代码，那么我的结果集将如下所示

name    | code  | total   |  date
build1    x1      10        04-2013
build1    x50     60        05-2013
build1    x1      80        06-2013
build1    x90     450       07-2013

我能够转置所有行，所以所有列都是月份，总数低于它。我的更新结果现在看起来像这样

name    |  code | apl |  may | jun | jul
build1     x1      10    0     80     0
build1     x50     0     60     0     0
build1     x90     0     0      0     450

上面的代码是我正在寻找的结果，但我现在要做的是按当月订购一切，然后从当月开始一年。

因此，如果当前月份是7月，那么我的结果集将按此顺序排序

name    |  code | jul |  jun | may | apl
build1     x1      0     80     0     10
build1     x50     0      0     60     0
build1     x90     450    0     0      0

我遇到的问题是我使用别名作为月份名称。你不能从别名中获取月份。另外，据我所知，别名是静态的，因此一旦你设置它们就无法改变。将月份作为列名称的唯一方法是从数据集中提取它。但是当我将行转置到列中时，我必须使用别名，因为我使用case语句来获取每个月的所有总计。

编辑：抱歉，postgresql版本是8.4，这是我到目前为止的查询

SELECT 


pname,
code,
SUM(totaljanurary)  AS "Janurary",
SUM(totalfebruary)  AS "February",
SUM(totalmarch)     AS "March",
SUM(totalapril)     AS "April",
SUM(totalmay)       AS "May",
SUM(totaljune)      AS "June",
SUM(totaljuly)      AS "July",
SUM(totalaugust)    AS "August",
SUM(totalseptember) AS "September",
SUM(totaloctober)   AS "October",
SUM(totalnovember)  AS "November",
SUM(totaldecember)  AS "December"

FROM(

SELECT 

    pname,
    code,
    SUM(case when extract (month FROM checked_date)=01 then total else 0 end) AS totaljanurary,
    SUM(case when extract (month FROM checked_date)=02 then total else 0 end) AS totalfebruary,
    SUM(case when extract (month FROM checked_date)=03 then total else 0 end) AS totalmarch,
    SUM(case when extract (month FROM checked_date)=04 then total else 0 end) AS totalapril,
    SUM(case when extract (month FROM checked_date)=05 then total else 0 end) AS totalmay,
    SUM(case when extract (month FROM checked_date)=06 then total else 0 end) AS totaljune,
    SUM(case when extract (month FROM checked_date)=07 then total else 0 end) AS totaljuly,
    SUM(case when extract (month FROM checked_date)=08 then total else 0 end) AS totalaugust,
    SUM(case when extract (month FROM checked_date)=09 then total else 0 end) AS totalseptember,
    SUM(case when extract (month FROM checked_date)=10 then total else 0 end) AS totaloctober,
    SUM(case when extract (month FROM checked_date)=11 then total else 0 end) AS totalnovember,
    SUM(case when extract (month FROM checked_date)=12 then total else 0 end) AS totaldecember



    FROM (

        --START HERE
        SELECT

            pname,
            code,
            COUNT(code)AS total,
            date_trunc('month',checked_date)::date AS checked_date


            FROM table1

            AND checked_date >= current_date-365
            AND checked_date <= current_date


            GROUP BY pname, code, date_trunc('month',checked_date)
    )T1
    GROUP BY pname, code, date_trunc('month',checked_date)

)T2
GROUP BY pname, code
ORDER BY pname, code

Answer 1

您希望crosstab()由附加模块tablefunc提供。

假设“date”的类型为date（就像它应该的那样）其他细节取决于您忘记提供的详细信息。

SELECT * FROM crosstab(
     $$SELECT name, code, to_char("date", 'mon'), total
       FROM   tbl
       WHERE  "date" <  now()
       AND    "date" >= now() - interval '1 year'
       ORDER  BY name, extract(month from now()) DESC$$

    ,$$VALUES
      ('dec'::text), ('nov'), ('oct'), ('sep'), ('aug'), ('jul')
    , ('jun'),       ('may'), ('apr'), ('mar'), ('feb'), ('jan')$$
   )
AS ct (name text, code text
   , dec int, nov int, oct int, sep int, aug int, jul int
   , jun int, may int, apr int, mar int, feb int, jan int);

有关其他说明，请参阅此密切相关的答案：
Sum by month and put months as columns

你不应该使用date作为标识符，它是一个保留字。我加倍引用它。
您不应该使用非描述性名称name。

将列名称显示为从current_date开始的过去一年的月份

1 个答案: