PHP - Strtotime

First time posting code and very new to php aka learning as I go...

The below code works as shown, but in the 'else' clause, I'd like to replace the +4 with
a variable.  How to do this?

I've tried,

Code: [Select]
$renewaldate = strtotime($subscriptiondate);

$sft = "' +" . $duesmultiplier . " year'";
$next_year = date('Y-m-d',strtotime($sft,$renewaldate));

but this doesn't return what I expect.  Perhaps this can't be done using this
technique?

Code: [Select]
if($n == 0)
 {
 $renewaldate = strtotime($subscriptiondate);
 $next_year = date('Y-m-d',strtotime('next year',$renewaldate));
 echo $next_year;
 }
else
 {
 $renewaldate = strtotime($subscriptiondate);
 $next_year = date('Y-m-d',strtotime('+4 year',$renewaldate));
 echo $next_year;
 }

Thank you

Hi!

the following code sucessfully outputs a timestamp.

Code: [Select]
$datealt=strtotime('2012-03-13 ,23:13:00');
The following however does not:

Code: [Select]
$datealt=strtotime('$date ,$hours:$minutes:00');

I have checked if the variables echo out and they do..

any clues much appreciated

x

so i have a basic html table and down the left column i have monday,tuesday,wednesday.... sunday.

In the column 2nd from the left, i want to fill it with the current weeks dates, which is really based on the current date. So if today is Jul 29th Friday, then strtotime("last monday") would give me Jul 26th Monday... my issue is "last monday" wont work if today is monday because that will give me 19th of july... and since I technically dont know what day is today i'm not sure how to cleanly work around this without having to use some if statement... if today is monday do "last monday +7 days" or just do "now" etc...

Thanks!

what is the propper way to write

Code: [Select]
$payrollweek="10";
date('Y-m-d', strtotime("+($payrollweek - 2) week $payrollend") ), "<br />";
its the ("+($payrollweek - 2)  that isnt working i think

how do I use strtotime() instead of mktime()?

$matchTimea = mktime($mHour, $mMinute, 0, $mMonth, $mDay, $mYear);

if($matchTimea-time() < $cutoffTimea*60)
{

I have this php code:

length is 15 , that is what it is coming out of the database


$length = $get['length'];

$time = date('Y-m-d', strtotime('+$length days'));

But when I echo time it comes out with this result 1969-12-31

What am I doing wrong?

Hi

I am new to PHP and trying to convert a date from a registration form (in the form 01/01/2011) to Y-m-d so that it can be stored in the database.

This is the code I have, pretty sure it worked before but not it has stopped working! Any ideas?

$dateformat = $_POST['dob'];
$correctformat = date('Y-m-d',strtotime($dateformat));

I have checked what $_POST['dob'] is printing and that is correct, however the $correctformat is printing 1970-01-01 everytime.

Any help would be greatly apreciated.

Thank you

Good Day,

I'm struggling to get strtotime() to return anything.

I have the following code..
$fields['birth']['day'] = 9;
        $fields['birth']['month'] = 7;
        $fields['birth']['year'] = 1982;
        $dob = $fields['birth']['day'].' '.$fields['birth']['month'].' '.$fields['birth']['year']; // Returns: 9 7 1982
        $dob = strtotime($dob); 
        die($dob);

No matter what I do $dob isn't returning anything once strtotime() is used. I've tried matching my day/month/year 's with examples in the manual on the strtotime() page but using date() on the fields, but it still doesn't want to return anything.

I feel I'm missing something simple here.

Any help is appreciated.

Regards,
Ace

hello,

what am i doing wrong?

Code: [Select]
$starttime=$row['starttime'];
$esttime1=date( 'g:i a', strtotime(-2 hours, $starttime) );

Hey freaks,

running into a little issue with the use of date and strtotime....

basically I wanna present the date in the format m-d-y to the user and insert it into the DB in the Y-m-d format. 

What seems to be happening is a freaky Y-d-m format and I don't know why this is.  This code is an example that expresses my frustration.

Code: [Select]
$nmon = strtotime("next Monday");
$next = date('m-d-Y', $nmon);
$_SESSION['REPORT_DUE'] = $next;


$nfri = strtotime("Friday");
$due = date('m-d-Y', $nfri);
$_SESSION['WEEK_ENDING'] = $due;
$ses1 = $_SESSION['WEEK_ENDING2'] = date('Y-m-d', strtotime($due));
$ses2 = $_SESSION['WEEK_ENDING3'] = date('d-m-Y', strtotime($ses1));


.....output....

Quote

[WEEK_ENDING] => 09-02-2011 [WEEK_ENDING2] => 2011-02-09 [WEEK_ENDING3] => 09-02-2011 )




Thank you for any assistance.

Little confuzzled.

I have got a bunch of listings that will be printed at the end of the page but I need to subtract a time specified in a textbox (send it to itself) on the page.

Code: [Select]
$row['active'] = (strtotime('$row[etd]') + $rwy1 = $_POST['tmatttextfield']);Well this just converts the time in the database which is always going to be a four character number such as: 1300 for 1PM into a time and adds the textbox value to the time.

Code: [Select]
<td width="50" bgcolor="<?php echo $bc?>"><strong><?php echo date('hi', $row['active']);?></strong></td>That just prints it all out formatting the time to only Hours and Minutes.

So just to clarify the user would put in say '10' in the textbox which would constitute 10 minutes.
The code above should then define that '10' as 10 minutes.
Define the database result, ie. 1500 as 3PM and add the 10 minutes to it so it would become 1510.

I have been getting some wierd and wonderful results whilst trying to get it to work. Mainly just 1200 though.

Harry.

can someone please tell me how to echo the current time or date?  I have the date in the format

1306768978

in the database but I dont want to echo it back that way obviously.   Not only that but does anybody know of a good tutorial on how to use this function?

Hi,
I'm struggling a bit with using a date format and strtotime()

Code: [Select]
if(!empty($_GET['utime']))
{
$time = strtotime($_GET['utime']);
}

utime is supplied in the format dd/mm/yyyy but when I use strtotime() it treats it as mm/dd/yyyy....so 07/12/2011 is taken as 12th Aug rather than 7th Dec..which is the date I want to use.

I had a quick ganders through the forums but couldn't find anything specific to this.

is there any other attributes to strtotime() that will help, or a similar fucntion or a neat way of swoping my dd/mm/yyyy round before i use strtotime()

any help greatly appreciated...

//
I am trying to calculate the number of days between dates and have a major problem.  When I try the example below, it gives me exactly 201 days, yet the dates have different times, meaning that a full day is impossible.  If it matters, I am testing via xampp 1.7.3 with php version 5.3.1 with Windows XP.  Can anyone explain what I am doing wrong.  Can daylight savings play a role?

        $frmritedate7116    =    strtotime('2010-12-13 23:00');
        $frmritedate7156    =    strtotime('2010-05-27 00:00');
//
        $frmritedate7251    =    $frmritedate7156 - $frmritedate7116;
//
//
1274943600-1292310000=-17366400 // 86400 seconds in a day ( 201.00 )
//
//
//

Hi - I have an HTML form where the customer can amend text dates. However, I have noticed that if you for example change 01 Jun 2012 to something like, 01 July 2012, 50% of the occasions, strtotime() will ignore the change and just go back to 01 jun 2012.

This is not unique to June, is there anything I can do to make things better ??

Thanks !