如何在SQL Server 2008中子字符串值

Question

我有以下字符串

KLPI_2012_CBBE2_E_12704_2012-09-21_13_59_52
IYT_2012_CBBI1_S_66_2012-09-21_15_28_53

我希望在第一次_和第五次之后提取所有内容_

例如:

2012_CBBE2_E_12704 
2012_CBBI1_S_66

Answer 1

方法1.嵌套的CHARINDEX调用.

使用获得第一的位置在.如果您将该位置作为第三个参数传递,您可以指示从某个位置开始搜索:CHARINDEX('_', stringvalue)_stringvalueCHARINDEX

CHARINDEX(`_`, stringvalue, startpos)

现在,如果起始位置将是结果,即如下:CHARINDEX('_', stringvalue)+1

CHARINDEX(`_`, stringvalue, CHARINDEX(`_`, stringvalue) + 1)

然后那会给你第二个_位置.所以,要找到第五个_,你需要再嵌套CHARINDEX三次:

WITH aTable AS (
  SELECT
    *
  FROM
    (VALUES
      ('KLPI_2012_CBBE2_E_12704_2012-09-21_13_59_52'),
      ('IYT_2012_CBBI1_S_66_2012-09-21_15_28_53')
    ) AS v (aStringColumn)
),
positions AS (
  SELECT
    aStringColumn,
    Underscore1 = CHARINDEX('_', aStringColumn),
    Underscore5 = CHARINDEX('_',
                            aStringColumn,
                            CHARINDEX('_',
                                      aStringColumn,
                                      CHARINDEX('_',
                                                aStringColumn,
                                                CHARINDEX('_',
                                                          aStringColumn,
                                                          CHARINDEX('_',
                                                                    aStringColumn
                                                                   ) + 1
                                                         ) + 1
                                               ) + 1
                                     ) + 1
                           )
  FROM
    aTable
)
SELECT
  aSubstring = SUBSTRING(aStringColumn,
                         Underscore1 + 1,
                         Underscore5 - Underscore1 - 1
                        )
FROM
  positions
;

方法2.字符拆分+排名.

1. 使用数字表将每个字符串拆分为单个字符,沿途拉出它们的位置.

2. 对字符串中每个字符的出现进行排名.

3. 获取两个子集:

   1)具有1的字符_和排名;

   2)具有5的性格_和排名.

   相互加入这些子集.

4. 类似于方法1,使用_#1和_#5 的相应位置来获取子串.

WITH aTable AS (
  SELECT
    *
  FROM
    (VALUES
      ('KLPI_2012_CBBE2_E_12704_2012-09-21_13_59_52'),
      ('IYT_2012_CBBI1_S_66_2012-09-21_15_28_53')
    ) AS v (aStringColumn)
),
split AS (
  SELECT
    t.aStringColumn,
    aChar    = SUBSTRING(t.aStringColumn, n.Number, 1),
    Position = n.Number
  FROM
    aTable t
    INNER JOIN Numbers n
      ON n.Number BETWEEN 1 AND LEN(t.aStringColumn)
),
ranked AS (
  SELECT
    *,
    rnk = ROW_NUMBER() OVER (PARTITION BY aStringColumn, aChar ORDER BY Position)
  FROM
    split
  WHERE
    aChar = '_'
)
SELECT
  aSubstring = SUBSTRING(first.aStringColumn,
                         first.Position + 1,
                         fifth.Position - first.Position - 1
                        )
FROM
  ranked first
  INNER JOIN ranked fifth
    ON first.aStringColumn = fifth.aStringColumn
WHERE
  first.rnk = 1
  AND fifth.rnk = 5
;

注意:两种方法都假设每个aStringColumn值都至少有5个下划线字符.

Answer 2

您可以使用SUBSTRING函数从给定字符串中提取子字符串.

SELECT SUBSTRING('KLPI_2012_CBBE2_E_12704_2012-09-21_13_59_52', 6, 18);