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Home/ Questions/Q 9082219
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T20:30:27+00:00 2026-06-16T20:30:27+00:00

In sql Server, I need to split a string based on number. What would

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In sql Server, I need to split a string based on number. What would be the best way for that.
For example, I need to split below string

  1. I am looking for this query. 2. Can you please help. 3. I would really appreciate that.

the result I am looking for is

  1. I am looking for this answer.
  2. Can you please help.
  3. I would really appreciate that.

Thanks.

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  1. Editorial Team
    Editorial Team
    2026-06-16T20:30:28+00:00Added an answer on June 16, 2026 at 8:30 pm

    Quick and dirty, but it works. Plenty of room for optimizing if you wish

    DECLARE @str AS VARCHAR(MAX);
    SET @str = '1. I am looking for this query. 2. Can you please help. 3. I would really appreciate that.';
    
    DECLARE @counter INT;
    DECLARE @table TABLE ([Text] VARCHAR(MAX));
    DECLARE @currentPattern VARCHAR(5);
    DECLARE @nextPattern VARCHAR(5);
    
    SET @counter = 1;
    WHILE 1=1
    BEGIN
        -- Set the current and next pattern to look for (ex. "1. ", "2. ", etc.)
        SET @currentPattern = '%' + CAST(@counter AS VARCHAR(4)) + '. %';
        SET @nextPattern = '%' + CAST(@counter + 1 AS VARCHAR(4)) + '. %';
    
        -- Check if the current pattern exists.
        IF (SELECT PATINDEX(@currentPattern, @str)) > 0
        BEGIN
            -- Check if the next pattern exists.
            IF (SELECT PATINDEX(@nextPattern, @str)) > 0
            BEGIN
                -- There is another pattern, so only get the text for the current one.
                INSERT INTO @table VALUES (SUBSTRING(@str, 1, PATINDEX(@nextPattern, @str) - 1));
                SET @str = SUBSTRING(@str, PATINDEX(@nextPattern, @str), LEN(@str) - PATINDEX(@nextPattern, @str) + 1);
            END
            ELSE
            BEGIN
                -- No other patterns exist, so just insert the variable text.
                INSERT INTO @table VALUES (@str);
            END
        END
        ELSE
        BEGIN
            -- Current pattern does not exist; break out of loop.
            BREAK;
        END
    
        SET @counter = @counter + 1;
    END
    SELECT * FROM @table;
    
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