获取每个ID的日期丢失的月份

时间：2019-01-05 06:36:59

标签： greenplum

我有一张包含会员编号和日期的表格。我想为每个成员显示从日期起缺少的月份。

这是输入表和期望表。

enter image description here

我尝试创建一个临时表来生成从最小月到最大月的序列，但是我无法将两个表连接起来。 谁能帮助我做到这一点。

到目前为止，这是我到目前为止尝试过的代码。

提取日期并将其存储在新列中

create temp table temp_table as select *, date_part('month', premiumpadidate) from training.premium  distributed by(memberid);

生成系列

create temp table temp_table_series as 
(select * from generate_series(cast((select min(date_part) from temp_table group by (memberid)) as integer), cast((select max(date_part) from temp_table group by (memberid)) as integer) )
) 
distributed by (generate_series)

我不明白如何将两个表合并在一起以获取每位员工的缺席月份。

请帮助我这样做

1 个答案:

答案 0 :(得分：0)

这很容易通过一个函数来完成，但是您可以一次选择完成。这是我为您提供的帮助。我用联合模拟了您的数据，并添加了一个“ 102”成员：

select 101 as id, '2016-01-01'::date as paidDate union
select 101 as id, '2016-02-01'::date as paidDate union
select 101 as id, '2016-04-01'::date as paidDate union
select 101 as id, '2016-08-01'::date as paidDate union
select 101 as id, '2016-11-01'::date as paidDate union
select 102 as id, '2016-01-01'::date as paidDate union
select 102 as id, '2016-02-01'::date as paidDate union
select 102 as id, '2016-04-01'::date as paidDate union
select 102 as id, '2016-08-01'::date as paidDate union
select 102 as id, '2016-11-01'::date as paidDate

您应该将它视为表格。

然后，为了检查月份，我在您的桌子上做了12个“案例”。如果找到月份，则输入“ 0”值，如果找不到月份，则输入月份值，如下所示：

select t.id, extract(month from t.paidDate),
    case when extract(month from t.paidDate) = 1 then 0 else 1 end as m1,
    case when extract(month from t.paidDate) = 2 then 0 else 2 end as m2,
    case when extract(month from t.paidDate) = 3 then 0 else 3 end as m3,
    case when extract(month from t.paidDate) = 4 then 0 else 4 end as m4,
    case when extract(month from t.paidDate) = 5 then 0 else 5 end as m5,
    case when extract(month from t.paidDate) = 6 then 0 else 6 end as m6,
    case when extract(month from t.paidDate) = 7 then 0 else 7 end as m7,
    case when extract(month from t.paidDate) = 8 then 0 else 8 end as m8,
    case when extract(month from t.paidDate) = 9 then 0 else 9 end as m9,
    case when extract(month from t.paidDate) = 10 then 0 else 10 end as m10,
    case when extract(month from t.paidDate) = 11 then 0 else 11 end as m11,
    case when extract(month from t.paidDate) = 12 then 0 else 12 end as m12
from (
    select 101 as id, '2016-01-01'::date as paidDate union
    select 101 as id, '2016-02-01'::date as paidDate union
    select 101 as id, '2016-04-01'::date as paidDate union
    select 101 as id, '2016-08-01'::date as paidDate union
    select 101 as id, '2016-11-01'::date as paidDate union
    select 102 as id, '2016-01-01'::date as paidDate union
    select 102 as id, '2016-02-01'::date as paidDate union
    select 102 as id, '2016-04-01'::date as paidDate union
    select 102 as id, '2016-08-01'::date as paidDate union
    select 102 as id, '2016-11-01'::date as paidDate
) t

此后，我们需要丢弃数据中存在的月份。我使用ID分组，在MIN月份开始。这将丢弃存在的月份（因为在这种情况下我们将它们归零了），并将维持不存在的月份。像这样：

select id, 
    min(m1) as m1, min(m2) as m2, min(m3) as m3, min(m4) as m4, min(m5) as m5, min(m6) as m6, 
    min(m7) as m7, min(m8) as m8, min(m9) as m9, min(m10) as m10, min(m11) as m11, min(m12) as m12
from 
(
    select t.id, extract(month from t.paidDate),
        case when extract(month from t.paidDate) = 1 then 0 else 1 end as m1,
        case when extract(month from t.paidDate) = 2 then 0 else 2 end as m2,
        case when extract(month from t.paidDate) = 3 then 0 else 3 end as m3,
        case when extract(month from t.paidDate) = 4 then 0 else 4 end as m4,
        case when extract(month from t.paidDate) = 5 then 0 else 5 end as m5,
        case when extract(month from t.paidDate) = 6 then 0 else 6 end as m6,
        case when extract(month from t.paidDate) = 7 then 0 else 7 end as m7,
        case when extract(month from t.paidDate) = 8 then 0 else 8 end as m8,
        case when extract(month from t.paidDate) = 9 then 0 else 9 end as m9,
        case when extract(month from t.paidDate) = 10 then 0 else 10 end as m10,
        case when extract(month from t.paidDate) = 11 then 0 else 11 end as m11,
        case when extract(month from t.paidDate) = 12 then 0 else 12 end as m12
    from 
    (
        select 101 as id, '2016-01-01'::date as paidDate union
        select 101 as id, '2016-02-01'::date as paidDate union
        select 101 as id, '2016-04-01'::date as paidDate union
        select 101 as id, '2016-08-01'::date as paidDate union
        select 101 as id, '2016-11-01'::date as paidDate union
        select 102 as id, '2016-01-01'::date as paidDate union
        select 102 as id, '2016-02-01'::date as paidDate union
        select 102 as id, '2016-04-01'::date as paidDate union
        select 102 as id, '2016-08-01'::date as paidDate union
        select 102 as id, '2016-11-01'::date as paidDate
    ) t
) t
group by t.id

您可以将执行任何规则作为结果，但是可以通过删除零来进一步清理它，如下所示：

select id,
    case when m1 = 0 then null else m1 end, case when m2 = 0 then null else m2 end,
    case when m3 = 0 then null else m3 end, case when m4 = 0 then null else m4 end,
    case when m5 = 0 then null else m5 end, case when m6 = 0 then null else m6 end,
    case when m7 = 0 then null else m7 end, case when m8 = 0 then null else m8 end,
    case when m9 = 0 then null else m9 end, case when m10 = 0 then null else m10 end,
    case when m11 = 0 then null else m11 end, case when m12 = 0 then null else m12 end
from 
(
    select id, 
        min(m1) as m1, min(m2) as m2, min(m3) as m3, min(m4) as m4, min(m5) as m5, min(m6) as m6, 
        min(m7) as m7, min(m8) as m8, min(m9) as m9, min(m10) as m10, min(m11) as m11, min(m12) as m12
    from 
    (
        select t.id, extract(month from t.paidDate),
            case when extract(month from t.paidDate) = 1 then 0 else 1 end as m1,
            case when extract(month from t.paidDate) = 2 then 0 else 2 end as m2,
            case when extract(month from t.paidDate) = 3 then 0 else 3 end as m3,
            case when extract(month from t.paidDate) = 4 then 0 else 4 end as m4,
            case when extract(month from t.paidDate) = 5 then 0 else 5 end as m5,
            case when extract(month from t.paidDate) = 6 then 0 else 6 end as m6,
            case when extract(month from t.paidDate) = 7 then 0 else 7 end as m7,
            case when extract(month from t.paidDate) = 8 then 0 else 8 end as m8,
            case when extract(month from t.paidDate) = 9 then 0 else 9 end as m9,
            case when extract(month from t.paidDate) = 10 then 0 else 10 end as m10,
            case when extract(month from t.paidDate) = 11 then 0 else 11 end as m11,
            case when extract(month from t.paidDate) = 12 then 0 else 12 end as m12
        from 
        (
            select 101 as id, '2016-01-01'::date as paidDate union
            select 101 as id, '2016-02-01'::date as paidDate union
            select 101 as id, '2016-04-01'::date as paidDate union
            select 101 as id, '2016-08-01'::date as paidDate union
            select 101 as id, '2016-11-01'::date as paidDate union
            select 102 as id, '2016-01-01'::date as paidDate union
            select 102 as id, '2016-02-01'::date as paidDate union
            select 102 as id, '2016-04-01'::date as paidDate union
            select 102 as id, '2016-08-01'::date as paidDate union
            select 102 as id, '2016-11-01'::date as paidDate
        ) t
    ) t
    group by 
    t.id
) t

就是这样，希望对您有所帮助。

致谢

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