Question

  select(sum(
       if (lower(st.c_gender) = 'male', 1, 0)) - ifnull(le.male, 0)) as male, (sum(
       if (lower(st.c_gender) = 'female', 1, 0)) - ifnull(le.female, 0)) as female, FROM_UNIXTIME(st.c_admissionDate, '%Y') as year, le.male, le.female
    from(SELECT s.*, zone.id as zoneid FROM groups s left join groups as zone ON s.parent_id = zone.id where s.id != s.parent_id) g, group_to_user u, student st
    left join(SELECT ifnull(sum(
       if (lower(l.c_gender) = 'male', 1, 0)), 0) as male, ifnull(sum(
       if (lower(l.c_gender) = 'female', 1, 0)), 0) as female, FROM_UNIXTIME(l.c_leavingDate, '%Y') as year from student l 
    where l.c_leavingDate is not null group by FROM_UNIXTIME(l.c_leavingDate, '%Y')) as le on le.year = FROM_UNIXTIME(st.c_admissionDate, '%Y') 
    where g.id = u.groupid and st.c_userId = u.userId group by FROM_UNIXTIME(st.c_admissionDate, '%Y')

以上查询返回以下输出

male    female  year    male    female
1860    657     1970    NULL    NULL
2        4      2012    NULL    NULL
3        0      2013    NULL    NULL
470     370     2014    0       1
0       365     2015    NULL    NULL
0       367     2016    NULL    NULL
1       260     2017    1       2

从上面的输出中需要将之前的值相加，如下面的输出

    male   female   year    male    female
    1860    657     1970    NULL    NULL
    1862    661     2012    NULL    NULL
    1865    661     2013    NULL    NULL
    2335    1031    2014    0       1
    2335    1396    2015    NULL    NULL
    2335    1763    2016    NULL    NULL
    2336    2023    2017    1       2

无论如何使输出如上所述帮助我

Answer 1

set @msum := 0;
set @fsum := 0;
select (@msum := @msum + T.male_sum) as male_sum, (@fsum := @fsum + T.female_sum) as female_sum, T.year, T.male, T.female
from (
select(sum(
       if (lower(st.c_gender) = 'male', 1, 0)) - ifnull(le.male, 0)) as male_sum, (sum(
       if (lower(st.c_gender) = 'female', 1, 0)) - ifnull(le.female, 0)) as female_sum, FROM_UNIXTIME(st.c_admissionDate, '%Y') as year, le.male, le.female
    from(SELECT s.*, zone.id as zoneid FROM groups s left join groups as zone ON s.parent_id = zone.id where s.id != s.parent_id) g, group_to_user u, student st
    left join(SELECT ifnull(sum(
       if (lower(l.c_gender) = 'male', 1, 0)), 0) as male, ifnull(sum(
       if (lower(l.c_gender) = 'female', 1, 0)), 0) as female, FROM_UNIXTIME(l.c_leavingDate, '%Y') as year from student l 
    where l.c_leavingDate is not null group by FROM_UNIXTIME(l.c_leavingDate, '%Y')) as le on le.year = FROM_UNIXTIME(st.c_admissionDate, '%Y') 
    where g.id = u.groupid and st.c_userId = u.userId group by FROM_UNIXTIME(st.c_admissionDate, '%Y')
)T;

简化此查询，但这会给您一个大致的想法。无法在非过程查询中完成。

Answer 2

您确实需要一个可靠排序的列。

但MySQL中的方法是使用像这样的@变量。

SELECT
      @male = @male + male        as male
    , @female = @female + female  as female
    , ordering_column
FROM (
       << your existing query here >>
     )
CROSS JOIN ( SELECT @male := 0 m, @female := 0 f ) vars
ORDER BY
     ordering_column

添加mysql查询的所有值的总和

2 个答案: