假设拥有数据集：
$p_name = "Berta";
$query = $articles->find()->contain([
'Comments' => function ($q) {
return $q
->select(['name'])
->where(['Comments.name' => $p_name]);
}
]);
我们发现与with
data_table(title, x) as (
select 'a', 1 from dual union all
select 'a', 3 from dual union all
select 'a', 4 from dual union all
select 'a', 5 from dual union all
select 'b', 1 from dual union all
select 'b', 2 from dual union all
select 'b', 3 from dual union all
select 'b', 6 from dual
)
select * from data_table;
TITLE | X
-----------
a 1
a 3
a 4
a 5
b 1
b 2
b 3
b 6
和a相关的点数不同。
如何对齐b列中的值，以便两个群组具有相同的点，填补X的空白？
预期结果是：
NULL我得到的直接解决方案是
TITLE | X
-----------
a 1
a NULL
a 3
a 4
a 5
a NULL
b 1
b 2
b 3
b NULL
b NULL
b 6
正如with
data_table(title, x) as (
select 'a', 1 from dual union all
select 'a', 3 from dual union all
select 'a', 4 from dual union all
select 'a', 5 from dual union all
select 'b', 1 from dual union all
select 'b', 2 from dual union all
select 'b', 3 from dual union all
select 'b', 6 from dual
),
all_points(x) AS (
select distinct x from data_table
),
all_titles(title) AS (
select distinct title from data_table
),
aligned_data(title, x) as (
select t.title, p.x from all_points p cross join all_titles t
)
select ad.title, dt.x
from aligned_data ad
left join data_table dt on dt.title = ad.title and dt.x = ad.x
order by ad.title, ad.x;
中cross join定义中的aligned_data是瓶颈，可能会破坏有价值数据集的性能。
我想知道这个任务是否可以更优雅地解决。也许可以提出一个带窗口函数的技巧。