[RESOLVED] preg_match() - need help with expression

steamPunk

I'm trying to write an expression that will test a date format so that it will validate a date string

basically i need to check that a date is in the yyyy-mm-dd format and that it isn't 0000-00-00

so far I've got this

"([0-9]{4})-([0-9]{2})-([0-9]{2})"

but it's wrong because I'm getting the error

Unknown modifier '-'

and also it will accept 0000-00-00 as ok - what i need is for it to say that in each part [0-9] is acceptable but it must have at least 1 instance of [1-9]

here's my code so far :

function testDate($dt){
	if(preg_match("([0-9]{4})-([0-9]{2})-([0-9]{2})",$dt,$regs)){
		echo $dt." is ok<br />";
	}else{
		echo $dt." is not ok<br />";	
	}
}

$str[1] = "2010-01-29";
$str[2] = "0000-00-00";
$str[3] = "0";
$str[4] = "";
$str[5] = "5876586868";

foreach($str as $v){
	testDate($v);
}

how can I make this work ?

I was tempted to just explode() the date string using "-" as a separator and then testing if the first four characters match "2009" or "2010" or "2010" etc but i would really like to understand how to do this with an expression

thanks

halojoy

One thing I see is you have no delimiter.
This is a character at the beginning and end of all expressions for preg_xxxx()

I use mostly # and #

if(preg_match("#([0-9]{4})-([0-9]{2})-([0-9]{2})#",$dt,$regs)){

steamPunk

thanks for your reply

ok - I've changed that and now i'm no longer getting an error message

but i still haven't figured out how to say that [0-9]{4} is accepted for the first four digits but at least one of the first four digits must contain only [1-9] - is such an expression possible ??!!

thanks

johanafm

halojoy;10940601 wrote:
One thing I see is you have no delimiter.

Actually he did have delimiters: ( and )
Although apparently unintentional, he did have pattern delimters: ( and )
Otherwise the error message would have been indicating lacking pattern delimiters.

To disallow 0000-00-00, just start with a negative look ahead assertion for that.

$p = '#(?!0000-00-00)...#';

But, you might as well use one assertion per group of digits, since neither of them can ever be 0, independently of the others:
There is no year 0
There is no month 0
There is no day 0

$p = '#(?!0000)...-(?!00)...-(?!00)...#';

steamPunk

thanks for your reply

I'd not seen the negative assertion thing before

however, when i do

function testDate($dt){
	if(preg_match('#(?!0000)...-(?!00)...-(?!00)...#',$dt,$regs)){
		echo $dt." is ok<br />";
	}else{
		echo $dt." is not ok<br />";	
	}
}

$str[1] = "2010-01-29";
$str[2] = "0000-00-00";
$str[3] = "0";
$str[4] = "";
$str[5] = "5876586868";

foreach($str as $v){
	testDate($v);
}

even the date 2010-01-29 doesn't match

so maybe I'm not doing this right ?

steamPunk

I've also tried this


'#([0-9]{4}(?!0000))-([0-9]{2}(?!00))-([0-9]{2}(?!00))#'

but it matches both '2010-01-09' and '0000-00-00' :/

paulnaj

Hi guys,

If you're doing this to learn about regexes then ignore this post.

If it's about parsing dates then have a look at a function from the comments in the manual http://www.php.net/manual/en/function.strtotime.php#94466 which will validate a date or return an empty string.

steamPunk

yes i was interested in learning more about regexes - but i'll have a look at that function as it may be useful in future projects

however, after much searching of the interwebs I found this :

'#[0-9]{4}-(((0[13578]|(10|12))-(0[1-9]|[1-2][0-9]|3[0-1]))|(02-(0[1-9]|[1-2][0-9]))|((0[469]|11)-(0[1-9]|[1-2][0-9]|30)))#'

and it seems to work ok !

thanks to all

johanafm

steamPunk;10940623 wrote:
I've also tried this
'#([0-9]{4}(?!0000))-([0-9]{2}(?!00))-([0-9]{2}(?!00))#'

My ... was to indicate stuff from your original pattern. In your code above you have put the lookAHEAD assertion after each part it's supposed to be compared against. Either move it to before where you expect it to show up, (?!0000)[0-9]{4}, or use a lookBEHIND assertion.