Question

I need to split a nvarchar(100) column into three nvarchar(28) columns without a known delimiter and without breaking mid-word. I am thinking that I would need to find the space that is near the 28th character, measure length and position of the word just before that space and decide weather the delimiter should be before or after that word. Then, again for the 3rd column.
From:

Col
-------------------------------------------------------------------------------------
Royal Mission Open Hutch with 4-Arch Doors, Plain Glass (with Glass Shelves Standard)

To:

Col1                      Col2                      Col3
------------------------  ------------------------- --------------------------
Royal Mission Open Hutch  4-Arch Doors, Plain Glass (with Glass Shelves Standa

Any Ideas?
Thanks nab

I am using SQL 2008

Answer 1

Using CROSS APPLYs, you could try something like this:

WITH data (col) AS (
  SELECT CAST('Royal Mission Open Hutch with 4-Arch Doors, Plain Glass (with Glass Shelves Standard)' AS nvarchar(100))
)
SELECT
  col1, col2, col3
FROM data
CROSS APPLY (
  SELECT
    NULLIF(30 - CHARINDEX(' ', REVERSE(LEFT(col, 29))), 0)
) AS x1 (first_space_pos)
CROSS APPLY (
  SELECT
    LEFT(col, ISNULL(first_space_pos, 28)),
    LTRIM(NULLIF(SUBSTRING(col, ISNULL(first_space_pos, 29), 999), ''))
) AS x2 (col1, col23)
CROSS APPLY (
  SELECT
    NULLIF(30 - CHARINDEX(' ', REVERSE(LEFT(col23, 29))), 0)
) AS x3 (second_space_pos)
CROSS APPLY (
  SELECT
    LEFT(col23, ISNULL(second_space_pos, 28)),
    LTRIM(NULLIF(SUBSTRING(col23, ISNULL(second_space_pos, 29), 28), ''))
) AS x4 (col2, col3)
;

The first CROSS APPLY searches for the last space in the first 29 characters of the big column, and the second one uses the found position to produce the first smaller column, col1 and return the rest of the string as col23.

The next two CROSS APPLYs do perform the same manipulations on col23, thus producing col2 and col3. The only difference is, the last CROSS APPLY puts into col3 at most 28 characters rather than all the remaining ones.

The hardcoded values like 28, 29 and 30 could be parametrised, but I’ll leave that part of the job to you.

You can try this query at SQL Fiddle.