Shell Programming and Scripting

Code:

case $Date in
  [0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]) echo 'DATE correct' ;;
  *) echo 'DATE incorrect' ;;
esac

Expanding on what Vino has already explained:

First, take a look on what your input COULD (legally) be. For instance, if your date format is "yyyymmdd":

"<year_number><1-12><1-31>"

Lets first stick with that and ignore for the moment, that February doesn't have 30+ days and that there are other dates too, which are illegal by that pattern: "20050431" for instance, because April has only 30 days.

Therefore our first shot at a regular expression matching dates could be:

Code:

 print - "$MyDateVar" | sed -n '/^[12][0-9]\{3\}[01][0-9][0123][0-9]/p'

if this prints anything it would be a date. The regexp matches:

a number "1" or "2", followed by three numbers 0-9 (the year), followed by
a number "0" or "1", followed by a number 0-9 (the month), followed by
a "0", "1", "2" or a "3", followed by a number 0-9 (the day).

Now for the year: right now we would match anything from 1000 to 2999. Could we further eliminate unwanted years? Yes! We will (presumably) have only to deal with dates of the last decade, so we could modify our regexp part for the year to:

Code:

[12][09][09][0-9]

This matches "1 or 2" followed by "0 or 9" followed by "0 or 9" followed by any digit. This still matches i.e. "2995", but it won't match "1875" or something such.

This way one could refine the regexp to even better spread the (syntactically) legal dates from the illegal ones. Upon request I will work out a complete solution.

bakunin

Code:

$ Date=20050629

$ touch -d $Date file1 && echo 'DATE correct' || echo 'DATE incorrect'
DATE correct

$ Date=20050631

$ touch -d $Date file1 && echo 'DATE correct' || echo 'DATE incorrect'
touch: invalid date format `20050631'
DATE incorrect

try

Code:

Date=19691231
Date=19700101
Date=20380119
Date=20380120

test on touch (coreutils) 5.0, correct date is from 1970/01/01 to 2038/01/19.