Uuups, seems I missed Johnny's suggestion yesterday - stopped reading when it seemed to be a format-cell solution only to me...
Malte. Johnny Rosenberg wrote, On 09/12/10 15:56: > Den 2010-09-12 14:19:27 skrev Malte Timmermann > <[email protected]>: > >> If you are sure it's always 10 chars long: >> >> Search: ([0-9][0-9][0-9])([0-9][0-9][0-9])([0-9][0-9][0-9][0-9]) >> Replace: $1-$2-$3 >> >> [0-9] means finding a digit >> () is for grouping, so you can use the found string in the replace >> statement >> $<n> is result of group <n> >> >> HTH, >> Malte. > > Or, if you didn't like my suggestion yesterday about > ([:digit:]{3})([:digit:]{3})([:digit:]{4}), then a mix of both suggestions > could be: > > ([0-9]{3})([0-9]{3})([0-9]{4}) > > Replace is the same in all three cases: > $1-$2-$3 > > {x} means x instances of what's to the left of it. > > So 3 digits followed by three digits followed by four digits will be > replaced by the first three digits followed by a minus sign and so on. > --------------------------------------------------------------------- To unsubscribe, e-mail: [email protected] For additional commands, e-mail: [email protected]
