Balazs Hegedus ha scritto:
$pattern =
'[0-3][0-9]/[0|1][0-9]/[1|2][0-9]{3,3}\s[0-2][0-9]:[0-5][0-9]';
I found this:
^([0-1]?\d)|(2[0-8]))\/((0?\d)|(1[0-2])))|(29\/((0?[1,3-9])|(1[0-2])))|(30\/((0?[1,3-9])|(1[0-2])))|(31\/((0?[13578])|(1[0-2]\/((19\d{2})|([2-9]\d{3}))|(29\/0?2\/((
Balazs Hegedus ha scritto:
I found this:
^([0-1]?\d)|(2[0-8]))\/((0?\d)|(1[0-2])))|(29\/((0?[1,3-9])|(1[0-2])))|(30\/((0?[1,3-9])|(1[0-2])))|(31\/((0?[13578])|(1[0-2]\/((19\d{2})|([2-9]\d{3}))|(29\/0?2\/[2468][048])|([3579][26]))00)|(((19)|([2-9]\d))(([2468]0)|([02468][48])|([1357
Oops, \s matches any whitespace character, so if you need only space
there you should change \s to space (this way it matches tab too).
Balazs
2006/3/31, Balazs Hegedus <[EMAIL PROTECTED]>:
> Hi,
>
> this regex isn't perfect at all but might do the job. You should
> modify the pattern to match th
Hi,
this regex isn't perfect at all but might do the job. You should
modify the pattern to match the year part against 2037 as a maximum
and also don't forget to checkdate().
Hope it helps,
Balazs
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Petar Nedyalkov ha scritto:
This is not correct since you don't check the ranges of the day digits, month
digits, etc.
The easiest way to check the string is to explode it by " " (space), then
explode the first part by slash and the second by column, and at last check
the ranges.
A regular exp
On Thursday 30 March 2006 16:49, Arie Nugraha wrote:
> try this :
>
> $date = "30/12/1982 15:30"
>
> if (preg_match("/\d{2}\/\d{2}\/\d{4}\s\d{2}:\d{2}/i",$date )) {
> echo "Date is valid";
> } else {
> echo "Date NOT valid";
> }
This is not correct since you don't check the ranges of the d
try this :
$date = "30/12/1982 15:30"
if (preg_match("/\d{2}\/\d{2}\/\d{4}\s\d{2}:\d{2}/i",$date )) {
echo "Date is valid";
} else {
echo "Date NOT valid";
}
hope it will help you
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Hi everybody,
do you have an regular expression example to check a string in a
"DD/MM/ hh:mm" format (without seconds)?
Bye and thanks,
Giacomo
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thanx a million!! you really saved my day!! :o)
hwee
- Original Message -
From: "John Holmes" <[EMAIL PROTECTED]>
To: "Ng Hwee Hwee" <[EMAIL PROTECTED]>
Cc: "PHP DB List" <[EMAIL PROTECTED]>
Sent: Thursday, August 05, 2004 11:05 AM
Subj
Ng Hwee Hwee wrote:
>>(sin|hkg|bkk|jkt)x[0-9]{4}-[0-9]{3}-[0-9]{2}
>
thanx!!! it worked like a charm! =)
just a question, i've tried adding a ^ to the front of your expressions and
a $ to the end of your expression and it still worked.. but is it
recommended? why didn't you use them?
^(sin|hkg|bkk|
al Message -
From: "John Holmes" <[EMAIL PROTECTED]>
To: "Ng Hwee Hwee" <[EMAIL PROTECTED]>
Cc: "PHP DB List" <[EMAIL PROTECTED]>
Sent: Thursday, August 05, 2004 10:22 AM
Subject: Re: [PHP-DB] Regular Expression
> Ng Hwee Hwee wrote:
>
>
Ng Hwee Hwee wrote:
Hi all,
I somehow just couldn't get my regular expression syntax correct. Will you please help
me?
some examples of my valid string is:
sinx0401-001-45
hkgx0403-020-12
jktx0402-000-01
bkkx0407-013-44
the definition is:
1st 4 characters - can only be "sinx" or "hkgx" or "bkkx"
Hi all,
I somehow just couldn't get my regular expression syntax correct. Will you please help
me?
some examples of my valid string is:
sinx0401-001-45
hkgx0403-020-12
jktx0402-000-01
bkkx0407-013-44
the definition is:
1st 4 characters - can only be "sinx" or "hkgx" or "bkkx" or "jktx"
next 4
[snip]
That does not work either... I can not make the statement fail!!!
I have tried
$password = "123456";
and
$password = "abcdef";
I have change the if statement to what is below and past in all letters
and
it still works?
if (preg_match ('/\d/', $password)) {
die ("You must have a numb
you only want to check to see if there is at least one digit?
If that is true, this should work:
if(preg_match('/\d/',$password)) {
echo 'password ok';
}
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hat there is at least 1 number in the password ?
// Larry
-Original Message-
From: Matthias Steinböck [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 24, 2004 1:41 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] regular expression help
hi!
is this correct: you want to check if there are
ok... this still does not do what you want, because it does not consider
that only digits should be between the letters... here is the correct
solution:
// dies
$password = 'alfgoesswimming';
// dies too
$password = 'a2lf4g2o4e7s9s3w9i0m5m7i0n3g';
// survives
$password = 'a2$5l5f4g2o4e7s9s3w9i0
hi!
is this correct: you want to check if there are two letters in the
password wich do not surround a digit? if so this is what you need:
// dies
$password = 'alfgoesswimming';
// dies too
$password = 'a2lf4g2o4e7s9s3w9i0m5m7i0n3g';
// survives
$password = 'a2l5f4g2o4e7s9s3w9i0m5m7i0n3g';
$foun
die ("You must have a number between 2 letters in your password ...
0-9");
// Larry
-Original Message-
From: Matt Matijevich [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 24, 2004 12:11 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] regular expression hel
try this
$password = 'abcdef';
if (preg_match ('/\w\d\w/', $password)) {
die ("You must have a number between 2 letters in your password ...
0-9");
}
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The *if* statement should fail with the password abcdef being qualified but
it never fails ?
If I understand the expression right, it should have at 1 number in it
between 2 letters ?
$password = abcdef;
if (preg_match ("/[A-z]+[0-9]+[A-z]+/", $password)) {
die ("You must have a n
Hi,
I would like to test whether a given sting contains only character, it should also
support spaces.
eg "Course Book String". I am able to test "CourseBookString".
Also I would like to know how can we avoid wild chars like *, or any other char apart
from alpha numeric char.
Thanks.
Hi!
does someone know how to find an exact word in a content with html tags ?
I'm using a regexp like this :
$searchRegEx = "'\b".$word_search."\b'msi";
to get boundary word results, but... if the searched word is like 'word or
word. or anything else, the
regular expression doesn't work..
if I
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