Use the following expression in order to create a derived column
(DT_STR,2,1252)"\r\n"
By default Data Type of string “\r\n” is DT_WSTR so here I am casting the data type to DT_STR
Knowledge.ToString()
Use the following expression in order to create a derived column
(DT_STR,2,1252)"\r\n"
By default Data Type of string “\r\n” is DT_WSTR so here I am casting the data type to DT_STR
Thanks, worked perfectly. I was looking to replace new line characters with ‘ ‘. The below worked.
LTRIM(RTRIM(REPLACE([Field],(DT_STR,2,1252)”\r\n”,” “)))
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