I am trying to change a string which may have a date inside e.g.
"This is the test string with 22/12/2012. 23/12/12 could anywhere in the string"
I need to change above string so that date are in the format d-m-y i.e.
"This is the test string with 22-12-2012. 23-12-12 could appear anywhere in the string"
EDIT:
Please note that the date will could changed in terms of years i.e. 2012 or 12 could be used at time i.e 20/06/2012, 20/06/12. Only year could be 2 or 4 digits, rest will be same.
Any help will be highly appreciated.
Cheers,
Use preg_replace like this:
$repl = preg_replace('~(\d{2})/(\d{2})/(\d{2,4})~', '$1-$2-$3', $str);
Live Demo: http://ideone.com/7HDNZa
$string = preg_replace("/([0-9]{2})\/([0-9]{2})\/([0-9]{2,4})/", "$1-$2-$3", $string);
The regex will find 3 lots of 2 numbers (or 2x2 + 1x4) separated by /'s and replace them with the same numbers separated by -'s.
You could try something like this:
preg_replace('~\b([0-2]?[1-9]|3[01])/(0?[1-9]|1[0-2])/(?=(?:\d\d|\d{4})\b)~', '$1-$2-', $str);
Should match valid dates only. Does match dates where the prefix 0 is not present, e.g. 4/16/13 if this is not desierable, remove the two first question marks (in [0-2]? and 0?)
Related
I've found this REGEXP:
^[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1])$
It's useful to validate a date in this format:
yyyy-mm-dd
with "-" as separator. It's perfect for me except that I'd like to make this:
1970-1-1
valid too, so I'm looking for a modification to have the zero optional in month and day block as first char. Could you please provide me some help?
If you want to have an optional zero, you can this RegEx (edited version of your own):
^[0-9]{4}-((?:0)?[1-9]|1[0-2])-((?:0)?[1-9]|[1-2][0-9]|3[0-1])$
To make a character ( or a range) optional in RegEx, put it in a parenthesis followed by a question mark:
[a-z] is mandatory
([a-z])? is optional, with capturing group
(?:[a-z])? without capturing group, just optional
Read more on Capturing group in RegEx
You could use something like so: ^[0-9]{4}-(([1-9])|(0[1-9]|1[0-2]))-(([1-9])|(0[1-9]|[1-2][0-9]|3[0-1]))$.
An example is available here.
Use your current regex just make the 0 optional 0?
^[0-9]{4}-(0?[1-9]|1[0-2])-(0?[1-9]|[1-2][0-9]|3[0-1])$
DEMO
PHP already has a construct to extract dates; if you're using a regex to extract the date instead, "Now you have two problems"
PHP's DateTime::createFromFormat: "Returns new DateTime object formatted according to the specified format"
You can use it's d or j option to accept 1 or 2 digit days and it's m or n option to accept 1 or 2 digit months.
The format you'll want to use is: "Y-m-d"
Im reluctant to ask but I cant figure out php preg_replace and ignore certain bits of the sting.
$string = '2012042410000102';
$string needs to look like _0424_102
The showing numbers are variable always changing and 2012 changes ever year
what I've tried:
^\d{4}[^\d{4}]10000[^\d{3}]$
^\d{4}[^\d]{4}10000[^\d]{3}$
Any help would be appreciated. I know it's a noob question but easy points for whoever helps.
Thanks
Your first regex is looking for:
The start of the string
Four digits (the year)
Any single character that is not a digit nor { or }
The number 10000
Any single character that is not a digit nor { or }
The end of the string
Your second regex is looking for:
The start of the string
Four digits (the year)
Any four characters that are not digits
The number 10000
Any three characters that are not digits
The end of the string
The regex you're looking for is:
^\d{4}(\d{4})10000(\d{3})$
And the replacement should be:
_$1_$2
This regex looks for:
The start of the string
Four digits (the year)
Capture four digits (the month and day)
The number 10000
Capture three digits (the 102 at the end in your example)
The end of the string
Try the following:
^\d{4}|10000(?=\d{3}$)
This will match either the first four digits in a string, or the string '10000' if there are three digits after '10000' before the end of the string.
You would use it like this:
preg_replace('/^\d{4}|10000(?=\d{3}$)/', '_', $string);
http://codepad.org/itTgEGo4
Just use simple string functions:
$string = '2012042410000102';
$new = '_'.str_replace('10000', '_', substr($string, 4));
http://codepad.org/elRSlCIP
If they're always in the same character locations, regular expressions seem unnecessary. You could use substrings to get the parts you want, like
sprintf('_%s_%s', substr($string,4,4), substr($string,13))
or
'_' . substr($string,4,4) . '_' . substr($string,13)
I have created a database of cigarette and trading cards. Each set title has a year associated with it. For example, 1943 or 2011. The year is always 4 characters long, but can be anywhere in the string.
Could someone please help me create a regex that will find the year in the string. I tried '/d{4}\b/' but it is failing.
(19|20)[0-9][0-9]
This will read in only 1900 and 2000 ranged dates.
Try this one :
/\b\d{4}\b/
it will match 4 digits embeded with non-words
d{4}\b will match four d's at a word boundary. You forgot the backslash in the character class: should be \d{4}\b. Depending on the input data you may also want to consider adding another word boundary (\b) at the beginning.
Here is a full solution:
$stringWithYear = '1990 New York Marathon';
$stringNoYear = preg_replace('/(19|20)[0-9][0-9]/', '', $stringWithYear);
echo trim($stringNoYear); // outputs 'New York Marathon'
This too works.. preg_match("/^1[0-9]{3}$/",$value))
Checks year of only starting with 1. You could change according to your requirement..
I am trying to extract a date from a string variable and was hoping to get some help.
$editdate = "Content last modified on 17 May 2011 at 23:13";
from this string, I am trying to extract 17 May 2011, please keep in mind that the date will vary and the code needs to be able to extract any date in this format, DD MMM YYYY.
I thought of using preg_match to do this but I couldn't come up with a proper regex pattern that would extract the date properly.
Is this possible to do with regex or should I use a different function?
Thanks for the help !
Try:
$timestamp = strtotime( str_replace( array("Content last modified ", "at"), "", $editdate ) );
Which will leave you with an epoch time stamp that you can then output however you like using date()
This is possible with a regex. Given the format DD MMM YYYY you would need a regex that matches two (or one?) digits, then one space, three letters, one space and four digits.
That would look like:
$regex = '/(\d{2} [a-z]{3} \d{4})/i';
This can be optimized further.
Presuming the textual content of your string is always the same, and that it always ends with the time...
$editdate = substr($editdate, 25, -9); // 17 May 2011
However, this is very inflexible if the date format were ever to change.
Try this 'un:
preg_match('/(\d?\d [A-Za-z]+ \d\d\d\d) at (\d\d\:\d\d)/', $editdate, $matches);
print_r($matches);
$date = $matches[1];
$time = $matches[2];
I THINK that'll work in all cases (though it is pretty ugly).... :)
This might be the pattern that does the trick:
([0-9]){1}([0-9]){0,1}(\s.*\s)([0-9]){4}
Search for 1 digit then there might be another, followed by a space and character, a space and 4 digits for the year.
I have textarea where users enter details, I want to extract year from the textarea. The problem is users enter year in different ways.
some users input dates as 1999 or '99 or september 99. how can I use a single regex string to extract year.
This should work for extracting the mentioned dates out of text:
preg_match_all('/(^|\s+)(\d{4}|\'\d{2}|(january|february|march|april|may|june|july|august|september|october|november|december) \d{2})(\s+|$)/i', $text, $matches);
This came to my mind, too (it gives the user a little more freedom):
preg_match_all('/(^|\s+)((january|february|march|april|may|june|july|august|september|october|november|december|jan|feb|mar|apr|may|jun|jul|aug|sep|oct|nov|dec)?(\s|\s?\')(\d{2}|\d{4}))(\s+|$)/i', $text, $matches);
All of the above would be easier when parsed with multiple expressions. Why do you need a single one?
If you simply want to parse a string that contains nothing else than this, you should use PHP's strtotime()-function.
\b\d\d(?:\d\d)?\b
will match a two- or four-digit number. So if your string only contains year numbers, that should be enough.
if (preg_match('/\b\d\d(?:\d\d)?\b/', $subject, $regs)) {
$result = $regs[0];
}
You can try to use strtotime:
strtotime — Parse about any English textual datetime description into a Unix timestamp
date('Y', strtotime($userinput));
[. -]((?:\d\d|')\d\d)$
Extract only digits from your string :
/\b(\d+)\b/