data extract

for example: I need  this string to be extracted from the above string :

1979 date sale.

SELECT
    ColumnNameHere,
 SUBSTRING(ColumnNameHere,
        ISNULL(2 + LEN(ColumnNameHere) - NULLIF(CHARINDEX('<B>', REVERSE(ColumnNameHere)), 0), 0),
        CASE CHARINDEX('</B>', ColumnNameHere) WHEN 0 THEN LEN(ColumnNameHere) + 1 ELSE CHARINDEX('</B>', ColumnNameHere) END -
        ISNULL(2 + LEN(ColumnNameHere) - NULLIF(CHARINDEX('<B>', REVERSE(ColumnNameHere)), 0), 0)) AS Comparison

Here is a working example

declare @x nvarchar(100)
select @x = '<B>1979 date sale.</B></I> NGC Census: (398/1885). PCGS Population (545/1407).'

select replace(left(@x, CHARINDEX('</B>', @x) -1), '<B>', '')

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Here is an example from a table

select replace(left(myColumn, CHARINDEX('</B>', myColumn) -1), '<B>', '') from myTable

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Hi Lee, thanks for the example, it says 'Invalid length parameter passed to the LEFT or SUBSTRING function', what could be the reason?

try this.

select CASE WHEN @x LIKE '%<B>%</B>%' 
            THEN SUBSTRING(@x, CHARINDEX('<B>',@x)+3,CHARINDEX('</B>',@x)-CHARINDEX('<B>',@x)-3) END

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declare @x nvarchar(100)
select @x = '<A>other</A><B>1979 date sale.</B></I> NGC Census: (398/1885). PCGS Population (545/1407).'

select CASE WHEN @x LIKE '%<B>%</B>%' 
            THEN SUBSTRING(@x, CHARINDEX('<B>',@x)+3,CHARINDEX('</B>',@x)-CHARINDEX('<B>',@x)-3) END
-- 1979 date sale.

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You probably have some null values in the column. Put a where clause in there to filter nulls

thanks