Sql Oracle-唯一Listagg值

我不熟悉使用Listag。下面的脚本在这方面起作用，它为我提供了一个值列表，但该列表重复了这些值

是否可以使用Listagg仅返回唯一的值列表

我正在使用oracle 10g

select distinct ds.catnr,ds.planqty, ds.ordnr, ds.posnr, ds.segnr, 
listagg(case when not li.paco is null then (select unique max(li1.paco) from leos_item li1 where li.av_part_no = li1.av_part_no) end, ', ') within group (order by pd.part_no) inq_no
from oes_delsegview ds, oes_address ad, oes_opos op, oes_nrbom nr, scm_prodtyp sp, leos_item li, part_description pd
where ds.delnr = ad.key
and ad.adr = ds.deladr
and ds.pos_o_status not in ('9', 'D')
and ds.pos_c_status not in ('9', 'D')
and ds.seg_o_status not in ('9', 'D')
and ds.seg_c_status not in ('9', 'D')
and ds.cunr in ('W31170','W31172')
and ds.pos_type != 'RC'
and ds.ordnr = op.ordnr
and ds.posnr = op.posnr
and ds.catnr = pd.catnr
and ds.prodtyp = pd.prodtyp
and ds.packtyp = pd.packtyp
and ds.catnr = nr.p_catnr (+)
and ds.prodtyp = nr.p_prodtyp (+)
and ds.packtyp = nr.p_packtyp (+)
and nr.c_prodtyp = sp.prodtyp (+) 
and sp.prodgrp (+) = 'COMP'
and substr(nr.c_prodtyp,1,2) not in ('MT','LF')
and nr.c_catnr = li.catnr (+)
and nr.c_prodtyp = li.prodtyp (+)
and nr.c_packtyp = li.packtyp (+)
and pd.catnr = '9780007938797'
group by ds.catnr,ds.planqty, ds.ordnr, ds.posnr, ds.segnr

我的列表的结果是：

14/061127-12, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16, 14/061127-16

我想看到的是：

14/061127-12, 14/061127-16

---

任何帮助都将不胜感激。

我删除了第一个

不同的
，因为您的
选择查询中的
分组依据
所有字段，并用
选择
查询替换了
时的
大小写：

select ds.catnr,ds.planqty, ds.ordnr, ds.posnr, ds.segnr, 
    listagg((select distinct max(li1.paco) from leos_item li1 where li.av_part_no = li1.av_part_no and li.paco is not null), ', ') within group (order by pd.part_no) inq_no
    from oes_delsegview ds, oes_address ad, oes_opos op, oes_nrbom nr, scm_prodtyp sp, leos_item li, part_description pd
    where ds.delnr = ad.key
    and ad.adr = ds.deladr
    and ds.pos_o_status not in ('9', 'D')
    and ds.pos_c_status not in ('9', 'D')
    and ds.seg_o_status not in ('9', 'D')
    and ds.seg_c_status not in ('9', 'D')
    and ds.cunr in ('W31170','W31172')
    and ds.pos_type != 'RC'
    and ds.ordnr = op.ordnr
    and ds.posnr = op.posnr
    and ds.catnr = pd.catnr
    and ds.prodtyp = pd.prodtyp
    and ds.packtyp = pd.packtyp
    and ds.catnr = nr.p_catnr (+)
    and ds.prodtyp = nr.p_prodtyp (+)
    and ds.packtyp = nr.p_packtyp (+)
    and nr.c_prodtyp = sp.prodtyp (+) 
    and sp.prodgrp (+) = 'COMP'
    and substr(nr.c_prodtyp,1,2) not in ('MT','LF')
    and nr.c_catnr = li.catnr (+)
    and nr.c_prodtyp = li.prodtyp (+)
    and nr.c_packtyp = li.packtyp (+)
    and pd.catnr = '9780007938797'
    group by ds.catnr,ds.planqty, ds.ordnr, ds.posnr, ds.segnr

我使用了一个regexp_replace函数来删除listagg中的重复项

regexp_replace(
    listagg((select distinct max(li1.paco) from leos_item li1 where li.av_part_no = li1.av_part_no and li.paco is not null), ', ') within group (order by pd.part_no) 
    ,'([^,]+)(,\1)+', '\1') inq_no

看起来工作正常

您可以通过正则表达式替换来完成。以下是一个例子：

-- Citations Per Year - Cited Publications main query. Includes list of unique associated core project numbers, ordered by core project number.
SELECT ptc.pmid AS pmid, ptc.pmc_id, ptc.pub_title AS pubtitle, ptc.author_list AS authorlist,
  ptc.pub_date AS pubdate,
  REGEXP_REPLACE( LISTAGG ( ppcc.admin_phs_org_code || 
    TO_CHAR(ppcc.serial_num,'FM000000'), ',') WITHIN GROUP (ORDER BY ppcc.admin_phs_org_code || 
    TO_CHAR(ppcc.serial_num,'FM000000')),
    '(^|,)(.+)(,\2)+', '\1\2')
  AS projectNum
FROM publication_total_citations ptc
  JOIN proj_paper_citation_counts ppcc
    ON ptc.pmid = ppcc.pmid
   AND ppcc.citation_year = 2013
  JOIN user_appls ua
    ON ppcc.admin_phs_org_code = ua.admin_phs_org_code
   AND ppcc.serial_num = ua.serial_num
   AND ua.login_id = 'EVANSF'
GROUP BY ptc.pmid, ptc.pmc_id, ptc.pub_title, ptc.author_list, ptc.pub_date
ORDER BY pmid;

超级简单的答案解决了

select
regexp_replace('14/061127-12, 14/061127-16, 14/061127-16','([^,]+)(,\1)*(,|$)', '\1\3')


from dual

->

14/061127-12，14/061127-16

查看我的完整答案
Hi haytem，谢谢，但是我仍然有相同的结果（重复值列表）。这个正则表达式替换有一个bug。尝试查看，搜索：替换为“查看搜索”。我将把固定的一个作为单独的答案发布。除非首先对字符串进行排序，否则的可能重复项不起作用。或者更确切地说，为了让这个正则表达式工作，列表中的所有副本必须彼此相邻。请参阅上面链接中的完整答案