替换SQL Server中的REGEXP_SUBSTR

时间:2016-03-23 22:15:39

标签: sql-server regex oracle

我正在寻找一种在SQL Server中替换使用INSTR(...)和REPLACE(REGEXP_SUBSTR(...))oracle函数的方法。

原始Oracle:

   SELECT 
      Name,
      CASE
           WHEN SUBSTR (NAME, 1, 2) = 'CG'
           THEN SUBSTR (NAME,INSTR (NAME,'_',1,2)+ 1,LENGTH (NAME))
           ELSE REPLACE (REGEXP_SUBSTR (NAME,'_[^_]+',1,2),'_','')
        END AS OPT,
        CASE
           WHEN SUBSTR (NAME, 1, 2) = 'CG'
           THEN SUBSTR (NAME,INSTR (NAME, '_',1,1) + 1, LENGTH (NAME))
           ELSE REPLACE (REGEXP_SUBSTR (NAME,'_[^_]+',1,1),'_','')
        END as Name2

已更新示例输出:

+---------------------------------+----------------------+------------------------------+
|              NAME               |         OPT          |            Name2             |
+---------------------------------+----------------------+------------------------------+
| AE 344592001H 6186694           | NULL                 | NULL                         |
| AE_161038002_6044777            | 6044777              | 161038002                    |
| BC_VIVS_HNB011A_1WAM            | HNB011A              | VIVS                         |
| BC_56230A_30SP                  | 30SP                 | 56230A                       |
| CG_3334902_NETWK_ ACTLM_3334912 | NETWK_ ACTLM_3334912 | 3334902_NETWK_ ACTLM_3334912 |
| CG_3334574_HMO1_CORACT_3334575  | HMO1_CORACT_3334575  | 3334574_HMO1_CORACT_3334575  |
| CG_3207160_POSC_1502AH_3207161  | POSC_1502AH_3207161  | 3207160_POSC_1502AH_3207161  |
| UH_141015_RHM                   | RHM                  | 141015                       |
| UH_127757_RIV                   | RIV                  | 127757                       |
| UH 523725 RIV                   | NULL                 | NULL                         |
| BS_W0055785_C500_M0005672       | C500                 | W0055785                     |
+---------------------------------+----------------------+------------------------------+

我试着查看charindex和patindex,但没有任何消息,我认为最初的正则表达式是我的头脑。关于如何在sql server中模仿这个逻辑的任何想法?

1 个答案:

答案 0 :(得分:1)

SQL Fiddle

MS SQL Server 2014架构设置

WITH Names ( lvl, name, remaining, idx ) AS (
  SELECT 1,
         name,
         name,
         CHARINDEX( '_', name )
  FROM   table_name
  UNION ALL
  SELECT lvl+1,
         name,
         SUBSTRING(remaining,idx+1,LEN(remaining)-idx),
         CASE WHEN CHARINDEX( '_', remaining, idx+1 ) = 0
              THEN 0
              ELSE CHARINDEX( '_', remaining, idx+1 ) - idx
              END
  FROM   Names
  WHERE  idx > 0
)
SELECT Name,
       MAX( CASE WHEN lvl = 3 AND ( Name LIKE 'CG%' OR idx = 0 ) THEN remaining
                 WHEN lvl = 3 THEN SUBSTRING( remaining, 1, idx - 1 )
                 END ) AS OPT,
       MAX( CASE WHEN lvl = 2 AND ( Name LIKE 'CG%' OR idx = 0 ) THEN remaining
                 WHEN lvl = 2 THEN SUBSTRING( remaining, 1, idx - 1 )
                 END ) AS Name2
FROM   Names
GROUP BY Name

查询1

|                            Name |                  OPT |                        Name2 |
|---------------------------------|----------------------|------------------------------|
|           AE 344592001H 6186694 |               (null) |                       (null) |
|            AE_161038002_6044777 |              6044777 |                    161038002 |
|                  BC_56230A_30SP |                 30SP |                       56230A |
|            BC_VIVS_HNB011A_1WAM |              HNB011A |                         VIVS |
|       BS_W0055785_C500_M0005672 |                 C500 |                     W0055785 |
|  CG_3207160_POSC_1502AH_3207161 |  POSC_1502AH_3207161 |  3207160_POSC_1502AH_3207161 |
|  CG_3334574_HMO1_CORACT_3334575 |  HMO1_CORACT_3334575 |  3334574_HMO1_CORACT_3334575 |
| CG_3334902_NETWK_ ACTLM_3334912 | NETWK_ ACTLM_3334912 | 3334902_NETWK_ ACTLM_3334912 |
|                   UH 523725 RIV |               (null) |                       (null) |
|                   UH_127757_RIV |                  RIV |                       127757 |
|                   UH_141015_RHM |                  RHM |                       141015 |

<强> Results

CHARINDEX( expressionToFind, expressionToSerach[, startIndex] )

查询2

_可用于查找单词中CHARINDEX( '_', name )的实例。

  • _会找到CHARINDEX( '_', name, CHARINDEX( '_', name ) + 1 )的第一个实例的索引。
  • _会找到0的第二个实例的索引,如果没有两个_字符,则会返回CHARINDEX( '_', name, CHARINDEX( '_', name, CHARINDEX( '_', name ) + 1 ) + 1)
  • _会找到0的第三个实例的索引,如果没有三个_个字符,则会返回SUBSTRING

将其嵌入内部选择中,您可以使用它在外部选择中获取相应的SELECT name, CASE WHEN idx2 > idx1 AND ( Name LIKE 'CG%' OR idx3 = 0 )THEN SUBSTRING( name, idx2 + 1, LEN( name ) ) WHEN idx3 > idx2 THEN SUBSTRING( name, idx2 + 1, idx3 - idx2 - 1 ) END AS OPT, CASE WHEN name LIKE 'CG%' THEN SUBSTRING( name, idx1 + 1, LEN( name ) ) WHEN idx2 > idx1 THEN SUBSTRING( name, idx1 + 1, idx2 - idx1 - 1 ) END AS Name2 FROM ( SELECT name, CHARINDEX( '_', name ) AS idx1, CHARINDEX( '_', name, CHARINDEX( '_', name ) + 1 ) AS idx2, CHARINDEX( '_', name, CHARINDEX( '_', name, CHARINDEX( '_', name ) + 1 ) + 1 ) AS idx3 FROM table_name ) t ,如下所示:

|                            name |                  OPT |                        Name2 |
|---------------------------------|----------------------|------------------------------|
|           AE 344592001H 6186694 |               (null) |                       (null) |
|            AE_161038002_6044777 |              6044777 |                    161038002 |
|            BC_VIVS_HNB011A_1WAM |              HNB011A |                         VIVS |
|                  BC_56230A_30SP |                 30SP |                       56230A |
| CG_3334902_NETWK_ ACTLM_3334912 | NETWK_ ACTLM_3334912 | 3334902_NETWK_ ACTLM_3334912 |
|  CG_3334574_HMO1_CORACT_3334575 |  HMO1_CORACT_3334575 |  3334574_HMO1_CORACT_3334575 |
|  CG_3207160_POSC_1502AH_3207161 |  POSC_1502AH_3207161 |  3207160_POSC_1502AH_3207161 |
|                   UH_141015_RHM |                  RHM |                       141015 |
|                   UH_127757_RIV |                  RIV |                       127757 |
|                   UH 523725 RIV |               (null) |                       (null) |
|       BS_W0055785_C500_M0005672 |                 C500 |                     W0055785 |

<强> Results

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