Re: [PHP-DB] date problem in MySQL DB

[CPT John W. Holmes]

From: Angelo Zanetti [EMAIL PROTECTED]

 This might be slightly off topic but hopefully someone can help.
 I have a field that is a varchar and I stored dates in it. But now I want
to
 change the type of the column to date, but I have a problem that the
formats
 differ:

 my format: mm/dd/
 mySQL format: -mm-dd

 So can I either:
 a. change the format of the date format

 or

 b. do i have to write a function that changes my format to the mySQL
before
 changing the column type??

You need to do option (b). Since this is a PHP list, I'd recommend
strtotime() and date() to do the formatting.

---John Holmes...

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RE: [PHP-DB] date problem in MySQL DB

[brett king]

Hi Angelo

Yes you will have to reformat and he is something that may help you.
Hope it does

function dateCheckMysql ($date) {

list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay))  (is_numeric($dateMonth)) 
(is_numeric($dateYear))) {

if (!checkdate($dateMonth, $dateDay, $dateYear)) {
return false;
} else {

return($dateYear./.$dateMonth./.$dateDay);
}
} else {

return false;
}
}

function dateCheckForm($date) {

list($dateYear, $dateMonth, $dateDay) = explode(-, $date);

if ((is_numeric($dateDay))  (is_numeric($dateMonth)) 
(is_numeric($dateYear))) {

return($dateDay./.$dateMonth./.$dateYear);
} else {
return false;
}
}

Many Thanks
Brett

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Re: [PHP-DB] date problem in MySQL DB

[Justin Patrin]

Brett King wrote:

Hi Angelo

Yes you will have to reformat and he is something that may help you.
Hope it does
	function dateCheckMysql ($date) {

list($dateDay, $dateMonth, $dateYear) = explode(/, $date);
if ((is_numeric($dateDay))  (is_numeric($dateMonth)) 
(is_numeric($dateYear))) {
if (!checkdate($dateMonth, $dateDay, $dateYear)) {
return false;
} else {
return($dateYear./.$dateMonth./.$dateDay);
}
} else {
return false;
}
}
	function dateCheckForm($date) {

		list($dateYear, $dateMonth, $dateDay) = explode(-, $date);

if ((is_numeric($dateDay))  (is_numeric($dateMonth)) 
(is_numeric($dateYear))) {
return($dateDay./.$dateMonth./.$dateYear);
} else {
return false;
}
}
Many Thanks
Brett
Wow, that's lots of code. I'd recommend something simpler. Like:

$mysqlDate = date('Y-m-d', strtotime($myDate));

$myDate = date('m/d/Y', strtotime($mysqlDate));

No checking involved, just one line of code.

--
paperCrane Justin Patrin
--
Question Everything, Reject Nothing
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