Following a Google…. Taking the code from the website: CREATE TABLE CRLF ( col1 VARCHAR(1000) ) INSERT CRLF SELECT ‘The quick brown@’ INSERT CRLF SELECT ‘fox @jumped’ INSERT CRLF SELECT ‘@over the ‘ INSERT CRLF SELECT ‘log@’ SELECT col1 FROM CRLF Returns: col1 —– The quick brown@ fox @jumped @over the log@ (4 row(s) affected) UPDATE CRLF SET col1 = REPLACE(col1, ‘@’, CHAR(13))
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