SELECT N rows before and after the row matching the condition?

Question

The behaviour I want to replicate is like grep with -A and -B flags . eg grep -A 2 -B 2 \"hello\" myfile.txt will give me all the lines wh

Answer 1

(MS SQL Server only)

The most reliable way would be to use the row_number function that way it doesn't matter if there are gaps in the id. This will also work if there are multiple occurances of the search result and properly return two above and below each result.

WITH

srt AS (
    SELECT ROW_NUMBER() OVER (ORDER BY id) AS int_row, [id]
    FROM theTable
),

result AS (
    SELECT int_row - 2 AS int_bottom, int_row + 2 AS int_top
    FROM theTable
        INNER JOIN srt
            ON theTable.id = srt.id
    WHERE ([message] like '%hello%')
)

SELECT theTable.[id], theTable.[message]
FROM theTable
    INNER JOIN srt
        ON theTable.id = srt.id
    INNER JOIN result
        ON srt.int_row >= result.int_bottom
        AND srt.int_row <= result.int_top
ORDER BY srt.int_row