Once upon a time, R. G. Newbury <newbury@xxxxxxxxxxxx> said: > # grep -h '[0-9]*s[0-9]*">' temp > Returns the example line with the 's[0-9]">' highlighted. A * in a regex is "0 or more of the previous", so basically you are just matching 's[0-9]*">' (because there will always be at least 0 of the [0-9] part at the start). If you really mean "1 or more", you can use an extended regex (the -E argument to grep/sed) and use + instead of *, so '[0-9]+s[0-9]*">'. -- Chris Adams <linux@xxxxxxxxxxx> _______________________________________________ users mailing list -- users@xxxxxxxxxxxxxxxxxxxxxxx To unsubscribe send an email to users-leave@xxxxxxxxxxxxxxxxxxxxxxx
