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What is the RegEx pattern for DateTime (2008-09-01 12:35:45 ) ?
I get this error:
No ending delimiter '^' found
Using:
preg_match('(?n:^(?=\d)((?<day>31(?!(.0?[2469]|11))|30(?!.0?2)|29(?(.0?2)(?=.{3,4}(1[6-9]|[2-9]\d)(0[48]|[2468][048]|[13579][26])|(16|[2468][048]|[3579][26])00))|0?[1-9]|1\d|2[0-8])(?<sep>[/.-])(?<month>0?[1-9]|1[012])\2(?<year>(1[6-9]|[2-9]\d)\d{2})(?:(?=\x20\d)\x20|$))?(?<time>((0?[1-9]|1[012])(:[0-5]\d){0,2}(?i:\ [AP]M))|([01]\d|2[0-3])(:[0-5]\d){1,2})?$)', '2008-09-01 12:35:45'); Gives this error:
Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 0 in E:\www\index.php on line 19
626
If you want to match for the actual '+', '. ' etc characters, add a backslash( \ ) before that character. This will tell the computer to treat the following character as a search character and consider it for matching pattern. Example : \d+[\+-x\*]\d+ will match patterns like "2+2" and "3*9" in "(2+2) * 3*9".
The regex matches on a date with the YYYY/MM/DD format and a "Date of birth:" or "Birthday:" prefix (Year min: 1900, Year max: 2020). For example: Date of birth: 1900/12/01. Date of birth: 2019.01.
Basically (0+1)* mathes any sequence of ones and zeroes. So, in your example (0+1)*1(0+1)* should match any sequence that has 1. It would not match 000 , but it would match 010 , 1 , 111 etc. (0+1) means 0 OR 1.
$ means "Match the end of the string" (the position after the last character in the string).
@Espo: I just have to say that regex is incredible. I'd hate to have to write the code that did something useful with the matches, such as if you wanted to actually find out what date and time the user typed.
It seems like Tom's solution would be more tenable, as it is about a zillion times simpler and with the addition of some parentheses you can easily get at the values the user typed:
(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2}) If you're using perl, then you can get the values out with something like this:
$year = $1; $month = $2; $day = $3; $hour = $4; $minute = $5; $second = $6; Other languages will have a similar capability. Note that you will need to make some minor mods to the regex if you want to accept values such as single-digit months.
178
A simple version that will work for the format mentioned, but not all the others as per @Espos:
(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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