Balazs Hegedus ha scritto:
?php
$date = '30/03/2983 12:00';
$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:
Balazs Hegedus ha scritto:
?php
$date = '30/03/2983 12:00';
$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:
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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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 day digits, month
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
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().
?php
$date = '30/03/2983 12:00';
$pattern = '[0-3][0-9]/[0|1][0-9]/[1|2][0-9]{3,3}\s[0-2][0-9]:[0-5][0-9]';
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 the
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 or
-
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:
Hi all,
I somehow just couldn't get my regular expression syntax correct
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?
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
Subject: Re: [PHP-DB] Regular Expression
Ng Hwee Hwee wrote
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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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 help
try this
$password = 'abcdef
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:
?php
// dies
$password = 'alfgoesswimming';
// dies too
$password = 'a2lf4g2o4e7s9s3w9i0m5m7i0n3g';
// survives
$password = 'a2l5f4g2o4e7s9s3w9i0m5m7i0n3g';
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:
?php
// dies
$password = 'alfgoesswimming';
// dies too
$password = 'a2lf4g2o4e7s9s3w9i0m5m7i0n3g';
// survives
$password =
in the password ?
// Larry
-Original Message-
From: Matthias Steinböck [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 24, 2004 1:41 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] regular expression help
hi!
is this correct: you want to check if there are two letters in the
password
you only want to check to see if there is at least one digit?
If that is true, this should work:
if(preg_match('/\d/',$password)) {
echo 'password ok';
}
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[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 number
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