Question

我坚持查询。我正在他们的笔记中搜索一个客户评论，但我需要它在之前，包括和在此查询找到的评论之后返回笔记：

SELECT ct.whn, ct.tpe, ct.drctn, ct.smmry, ct.dscrptn, ct.zxcrtdby,
(SELECT nme FROM sdcrdtr WHERE sdcrdtr.id = ct.crdtr) as 'Creditor Name',
(SELECT accntrfrnce FROM cddbt WHERE cddbt.id = ct.dbt) as Reference

FROM cdcntct ct, cdcse c

WHERE c.id = ct.cse AND c.rfrnce = '123456789'
AND ct.smmry LIKE '%review%'

我在这里看到过类似的问题，但它是用于从提供特定ID或类似内容的表中查找所有内容。

我找不到从哪里开始！

提前致谢。将

编辑：示例日期

ID    WhN         SMMRY            DSCRPTN
12    01/05/2015  Sent Paperwork   Info Received
13    01/05/2014  Reviewed         Some Info Received
14    28/04/2014  Spoke to Client  Said hello
15    01/05/2013  Note 15          Blank
16    01/05/2012  Note 16          Blank
17    01/05/2011  Review done      Blank
18    30/12/2010  Another note     Blank

所以如果我要运行我的查询，但是在之前，之后，之后的行请求，它将返回ID的12,13,14然后16,17,18。虽然它目前只返回ID 13和17，因为我不知道在这个条款中添加的位置。

Answer 1

你可以这样做：

select t2.* 
from MyTable t2
inner join (
  select t.ID, max(tb.ID) as ID_Before, min(ta.ID) as ID_After
  from MyTable t
  left join MyTable tb on tb.ID < t.ID
  left join MyTable ta on ta.ID > t.ID
  where t.SMMRY like '%review%'
  group by t.ID
) a on t2.ID in (a.ID, a.ID_Before, a.ID_After)

SQL Fiddle

| ID |        WhN |           SMMRY |            DSCRPTN |
|----|------------|-----------------|--------------------|
| 12 | 01/05/2015 |  Sent Paperwork |      Info Received |
| 13 | 01/05/2014 |        Reviewed | Some Info Received |
| 14 | 28/04/2014 | Spoke to Client |         Said hello |
| 16 | 01/05/2012 |         Note 16 |              Blank |
| 17 | 01/05/2011 |     Review done |              Blank |
| 18 | 30/12/2010 |    Another note |              Blank |

Mysql在多行之前和之后

1 个答案: