Replace first occurrence of substring in a string in SQL

You can do a CHARINDEX or a PATINDEX, as shown above, but I would also recommend adding a COALESCE, in case your @stringtoFind it not included in your @stringhere.

SELECT COALESCE(STUFF(@stringhere, PATINDEX('%' + @stringtofind + '%', @stringhere), LEN(@stringtofind), ' '), @stringhere)

I had the same problem and made the same response as Tim Biegeleisen, but in a function:

CREATE FUNCTION DBO.FN_REPLACE_FIRST(@X NVARCHAR(MAX), @F NVARCHAR(MAX), @R NVARCHAR(MAX)) RETURNS NVARCHAR(MAX) AS BEGIN
RETURN STUFF(@X, CHARINDEX(@F, @X), LEN(@F), @R)
END

So I just call the function instead:

SELECT DBO.FN_REPLACE_FIRST('Text example', 'ex', 'eexx') --> Returns 'Teexxt example'

The explanation is the same

it seems you miss 2% preceding and trailing to the target string

please try:

select STUFF(@stringhere, PATINDEX('%' + @stringtofind + '%', @stringhere), LEN(@stringtofind), ' ')

You can use a combination of STUFF and CHARINDEX to achieve what you want:

SELECT STUFF(col, CHARINDEX('substring', col), LEN('substring'), 'replacement')
FROM #temp

CHARINDEX('substring', col) will return the index of the first occurrence of 'substring' in the column. STUFF then replaces this occurrence with 'replacement'.