Question

I have combined two different tables together, one side is named DynDom and the other is CATH. I am trying to remove duplicates from that table such as below: enter image description here

However, if i select distinct Dyndom pdbcode from the table, it returns distinct values of that pdbcode. DynDom table and

CATH

根据以上情况,我对表格中的DynDom/CATH栏进行了评论,并对DynDom/CATH进行了单独处理,并相应地将这些数值退回,这是我需要的东西,我很想知道,我是否可以使用2个不同的说明,以纸浆编码为基础,将整个表格的不同价值回去。

我的法典如下:

select DISTINCT
    cath_dyndom_table_2."DYNDOM_DOMAINID",
    cath_dyndom_table_2."DYNDOM_DSTART",
    cath_dyndom_table_2."DYNDOM_DEND",
    cath_dyndom_table_2."DYNDOM_CONFORMERID",
    cath_dyndom_table_2.pdbcode,
    cath_dyndom_table_2."DYNDOM_ChainID",
    cath_dyndom_table_2.cath_pdbcode,
    cath_dyndom_table_2."CATH_BEGIN",
    cath_dyndom_table_2."CATH_END"
from 
    cath_dyndom_table_2 
where 
    pdbcode =  2hun 
order by 
    cath_dyndom_table_2."DYNDOM_DOMAINID",
    cath_dyndom_table_2."DYNDOM_DSTART",
    cath_dyndom_table_2."DYNDOM_DEND",
    cath_dyndom_table_2.pdbcode,
    cath_dyndom_table_2.cath_pdbcode,
    cath_dyndom_table_2."CATH_BEGIN",
    cath_dyndom_table_2."CATH_END";

最后,我要根据制版编码,从DynDom和CATH处搜索田地,并在没有重复值的情况下将电网退回。

谢谢。

UPDATE :

这是我已做的要务表。

    CREATE VIEW cath_dyndom_table AS
SELECT
  r.domainid AS "DYNDOM_DOMAINID",
  r.DomainStart AS "DYNDOM_DSTART",
  r.Domain_End AS "DYNDOM_DEND",
  r.ddid AS "DYN_DDID",
  r.confid AS "DYNDOM_CONFORMERID",
  r.pdbcode,
  r.chainid AS "DYNDOM_ChainID",
  d.cath_pdbcode,
  d.cathbegin AS "CATH_BEGIN",
  d.cathend AS "CATH_END"
FROM dyndom_domain_table r
  FULL OUTER JOIN cath_domains d ON d.cath_pdbcode::character(4) = r.pdbcode 
  ORDER BY confid ASC;

Answer 1

如果你希望从每个桌上就域名和地域范围达成联合国决定,那么就能够做到:

SELECT DYNDOM_DOMAINID, DYNDOM_DSTART, DYNDOM_DEND
FROM DynDom
UNION
SELECT RTRIM(cath_pdbcode), CATH_BEGIN, CATH_END
FROM CATH

这应当消除准确的重复(例如,域名、起步和终点完全相同),但不会消除不同范围重复的域名——如果存在的话,你需要决定如何处理这些域名(将它们作为单独条目加以保留,将其与最低的起步和最高终点相结合,或选择任何其他办法)。

EDIT:实际上,我认为你能够仅仅通过改变日本宇宙开发网的条件,取得预期结果,以便:

FULL OUTER JOIN cath_domains d 
ON d.cath_pdbcode::character(5) = r.pdbcode || r.chainid AND
   r.DomainStart <= d.cathbegin AND
   r.Domain_End >= d.cathend

Answer 2

你们正在做的是“两个表格”的<条码>。

为使一条没有重复的线路,有必要在<代码>1-to-1链接上在tables之间架设<>。

参看HERE,什么是cartesian并入和 rel=“nofollow”HERE 如何避免!

UPDATE :

友情链接