Richard,
You would need a (CHAR( )) value for blank space achieve your task.
Using The Glorious R:BASE 2000 (ver 6.5++) for Windows, here is
an example to take care of that.
Example 01:
SET VAR vValueList TEXT = NULL
SELECT (LISTOF(ColumnName)) INTO vValueList INDIC IvValueList +
FROM TableName WHERE ...
The variable vValueList will be a text string of the values separated by the current comma
delimiter character. If you would like to add a single space after each value, then:
SET VAR vValueList TEXT = NULL
SELECT (SRPL(LISTOF(ColumnName),',',', ',0)) INTO +
vValueList INDIC IvValueList FROM TableName WHERE ...
Notice the additional space after comma in ReplaceString.
If you would like to use a carriage return after each value, then:
SET VAR vValueList TEXT = NULL
SELECT (SRPL(LISTOF(ColumnName),',',(CHAR(10)),0)) INTO +
vValueList INDIC IvValueList FROM TableName WHERE ...
Hope that helps!
Enjoy and make sure to have fun!
Very Best Regards,
Razzak.
At 03:25 PM 5/21/2002 -0400, Rich ( RINGO) Starkey wrote:
How would you go about replacing a '.' without using the space ' ' in the
(SRPL(0)) function.
I have Imported Data in the table that looks like this:
Mr. & Mrs. A.
George C. & Linda V.
I want it to look like this:
Mr & Mrs A--- No punctuation and no additional spaces
George C & Linda V--- No punctuation and no additional spaces
Following is the statement I used:
Update Person set FirstName = (SRPL(FirstName,'.','',0)) Where FirstName
contains '.'
I receive the following error message " -Error- Argument 3 of function SRPL
cannot be INTEGER. (2153)" Third argument is two single quotes.
When I use the following statement the function works, however I have an
additional space in the data. Third argument no has a space between the
single quotes.
Update Person set FirstName = (SRPL(FirstName,'.',' ',0)) Where FirstName
contains '.' There is an extra space between Mr,Mrs and A
I cannot have a double space between the characters. (Mr__& Mrs__ A_)
