Working with 2014-06-14-1506p here is a helpful Regexp to reorder it

Using a most excellent tool from (den4b called Renamer) - one of the few tools I happily purchased and want to promote it!

The regular expression to rearrange the date (helpful for date-stamping files) is:


then it can be easily reordered:

For those of you unfamiliar with REGEX or ReGularEXpressions- it breaks down like this:

  • (.*)  =  $1
    • will match zero or more of any characters
    • By enclosing it in parenthesis () it becomes a sub expression. 
    • Since it is the first subexpression it can be referenced to be reordered with $1 in Renamer.
  • (   =  left parenthesis start of the second expression  $2  for the year-month-day
    • left parenthesis is the the start of the next expression to Capture. If the parenthesis weren't there the characters would match and be lost
    • [0-9]{4}  for the year 0000 through 9999
      • Square brackets [ ] enclose a set of characters for reference. 
      • 0-9 inside the brackets species a range of all characters 0 through 9
      • {4} Curled brackets indicate the number of characters to match in the previous expression.
      • By combining [0-9]{4} means 4 numbers, zero through nine- perfect for a date 2014
      • alternatively ([1-2]{1}[0-9]{3}) should cover the last century 1000 through 29
    • [-_]{0,1} is for a literal dash or underscore character - zero or one of
    • [0-1][0-9] for the month 
      • alternatively [0-1]{0,1}[0-9]  would allow for 06 or 6 for the month 

    • [-_]{0,1} is for a literal dash or underscore character
    • [0-3][0-9] for the day 01 through 39
      • alternatively [0-3]{0,1}[0-9]  would allow for a single digit day
  • ) =  right parenthesis END of the second expression  $2  for the year-month-day
  • [-_]{0,1} is for a literal dash or underscore character - zero or one of
  • (   =  left parenthesis start of the third expression  $3 
    • [0-2][0-9][0-5][0-9] = for time 0000 through 2959 (Covers through 2399 24 hour clock)
    • [am,a,p,pm]{0,1}  = zero or one of a, am, p, pm for  am pm--There is a glitch here I cannot figure out how to extend so it has the matching for am|a|pm|p - since the letters are the same it will not match the characters
  • ){0,1} = Right parenthesis END of the third expression $3 {0,1} means zero or one of these can exist
  • [-_]{0,1} is for a literal dash or underscore character - zero or one of
  • (.*) = $4 the 4th expression- the remainder (zero or more) of the remaining characters- in case the match is in front of the expression


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