选择具有相同值的多个行

Question

我有一张桌子,有点像这样:

ID  |  Chromosome | Locus | Symbol | Dominance |
===============================================
1   |      10     |   2   |   A    |   Full    |
2   |      10     |   2   |   a    |   Rec.    |
3   |      10     |   3   |   B    |   Full    |
4   |      10     |   3   |   b    |   Rec.    |

我想选择具有相同基因座和染色体的所有行.例如,第3行和第4行.一次可能超过2个,它们可能不是有序的.

我试过这个:

SELECT *
FROM Genes
GROUP BY Locus
HAVING Locus='3' AND Chromosome='10'

但它总是返回第3行,从不返回第4行,即使重复也是如此.我想我错过了一些明显而简单的东西,但我很茫然.

有人可以帮忙吗？

Answer 1

您需要了解当您GROUP BY在查询中包含时,您要告诉SQL组合行.每个唯一Locus值将获得一行.在Having随后过滤这些组.通常在选择列表中指定aggergate函数,如:

--show how many of each Locus there is
SELECT COUNT(*),Locus FROM Genes GROUP BY Locus

--only show the groups that have more than one row in them
SELECT COUNT(*),Locus FROM Genes GROUP BY Locus HAVING COUNT(*)>1

--to just display all the rows for your condition, don't use GROUP BY or HAVING
SELECT * FROM Genes WHERE Locus = '3' AND Chromosome = '10'

Answer 2

假设您希望所有行与另一行完全相同,Chromosome并且Locus:

您可以通过将表连接到自身来实现此目的,但只能从连接的一个"侧"返回列.

诀窍是将连接条件设置为"相同的基因座和染色体":

select left.*
from Genes left
inner join Genes right
on left.Locus = right.Locus and 
  left.Chromosome = right.Chromosome and left.ID != right.ID

您还可以通过在where-clause中添加过滤器来轻松扩展它.