使用CHARINDEX(‘_’,stringvalue)获取stringvalue中第一个_的位置.如果您将该位置作为第三个参数传递,则可以指示CHARINDEX从某个位置开始搜索：

CHARINDEX(`_`,stringvalue,startpos)

现在,如果起始位置是CHARINDEX(‘_’,stringvalue)1的结果,即如下：

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

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

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
                                                                   ) + 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.

>使用numbers table将每个字符串拆分为单个字符,沿途拉出它们的位置.
>对字符串中每个字符的出现进行排名.
>获取两个子集：

1)用字符_和排名1;

2)用字符_和排名5.

相互加入这些子集.
>与方法1类似,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'),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：两种方法都假定每个aStringColumn值至少有5个下划线字符.