RegEx expression that will match Date format

Question

I have a line with "1999-08-16"^^xsd:date. What will be the regex to capture the whole string as "1999-08-16"^^xsd:datein flex file? And, is it possible to capture only "1999-08-16" as a string. If yes, then what will be the regex for it in flex?

Answer 1

Try this one:

^"(\d{4}-(?:0?[1-9]|1[012])-(?:30|31|[12]\d|0?[1-9]))"\^\^xsd:date$

Explanation:

• ^ - start of line
• \d{4} - year part
• (?:0?[1-9]|1[012]) - month, can be:
  • 01-09 or 1-9 (that's why 0?)
  • 10,11,12 (1[012] part)
• ?: means non-capturing group (if we need just alternating matches with |, but not outputting them to user)
• (?:30|31|[12]\d|0?[1-9]) - day part, can be:
  • 30,31,
  • 10-29 (part [12]\d)
  • 1-9, or 01-09 (0?[1-9])
    Also we use non-capturing group for matching day
• $ matches end of line
• Everything in between "" is captured to standard capturing group, as you need to extract date

Demo

NOTICE: When matching days we put 1-9 day numbers in LAST alternating group:

(?:30|31|[12]\d|0?[1-9])

that's because regex engine when given alternating matches uses FIRST matched result and other matched alternatives are ignored. For example- in string 1 11 expression:

• (?:\d{2}|\d) gives 2 matches
• (?:\d|\d{2}) gives 3 matches

Answer 2

To capture whole string \"[0-9]{4}-[0-9]{2}-[0-9]{2}\"\^\^[^ ]* can be used.