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PHP - Using Strtotime To Add To A Time
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. Similar Tutorialshi i have a mysql table with upddate and updtime field as follows: upddate updtime 2011-02-25 11:03:05 2011-01-28 08:01:09 2011-02-20 07:00:08 2011-02-15 06:10:02 2011-02-05 10:21:04 2011-02-09 10:20:25 the data types for upddate is mysql date and for updtime its mysql time In my php page i use Code: [Select] <?php echo $date = date("d M Y",strtotime($row_rs1['upddate'])); ?> to retrieve date and the results displays as 25 Feb 2011. And for time right now im using Code: [Select] <?php echo $row_rs1['updtime']; ?> which displays time as 11:03:05 I want to display time as HH:MM only, how can i do it? I apologizing for bothering the expert here again, but I've run into another issue. I am using the following code because I can't change the default timezone on my shared hosting plan:
$time = date("h:i A", strtotime("+19 hours"));
there are few issues with this. if I just use: date("h:i: A") the time prints as 7 hours earlier than my timezone, and currently it shows 1:09 AM and it appears in the database as "01:09:00". if I use the code above, the time prints out as "08:09 PM", but it is inserted into the database as "08:09:00". Shouldn't it read "20:09:00" instead? and why isn't the time being inserted into the database with the "PM" suffix on it like it appears in the printout? the "time" field in my database is set to type TIME. thanks. Edited October 14, 2019 by ajetrumpetI'm trying to add 5 days to $today date: Code: [Select] <?php $exp_date = "2011-09-11"; $todays_date = date("Y-m-d"); $today = strtotime("+5 days", $todays_date); $expiration_date = strtotime($exp_date); if ($expiration_date > $today) { echo "Valid: Yes"; } else { echo "Valid: No"; } ?> Is +5 days used wrong? Hi, I have a bit of an issue with my application, I didn't notice until today, Sunday... Apparently PHP sees Sunday as the first day of the week, which I cannot fathom where it got hat notion from... but anyway, so I have the following samples of my code: Code: [Select] $mondayoflastweek = date('Y-m-d H:i:s', strtotime('monday last week')); $sundayoflastweek = date('Y-m-d 23:59:59', strtotime('sunday last week')); The crazy and very troublesome thing about the results of that code is that if you run that today, Sunday, it shows today as last week... So therefore, the next bit of code is just as problematic: Code: [Select] $mondayofthisweek = date('Y-m-d H:i:s', strtotime('monday this week')); $sundayofthisweek = date('Y-m-d 23:59:59', strtotime('sunday this week')); As you can imagine, for payroll, this is causing a big problem... My week starts at precisely 12:00 am on Monday and ends at precisely 11:59 pm on Sunday... Could someone please help me out with this... Hello! A user is going to input a date in the form "mm/dd/yyyy". I'd like to convert this to a time stamp so that I can store it via MySql. I've heard about the "strtotime" function in PHP but I'm not sure how it can tell that the user 03/04/2011 represents March,4th 2011 and not, say April,3rd 2011 (European version). Help with the correct syntax for conversion would be greatly appreciated. Thank you, Eric 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 So, my problem is that I need to edit my Xmas calculator to understand when this year's Xmas is over, it will automatically jump to the next one (2011-12-25). I have no idea how to do this (noob to php...) Thanks in advance. Here's my code: Code: [Select] <?php $time=time(); $xmas=strtotime("2010-12-25 00:00:00"); $diff = $xmas - $time; $days=intval($diff/86400); $left=$diff%86400; $hs=intval($left/3600); $left=$left%3600; $mins=intval($left/60); $secs=$left%60; echo "<font size=12 face=corbel> Xmas is after:<br> <strong>$days days</strong><br> <strong>$hs hours</strong><br> <strong>$mins mins $secs secs</strong>! </font>"; ?> Hello,
I'm trying to use strtotime to add time to a mysql type timestamp. Any help would be appreciated. I think I have tunnel vision from looking at it when I know it has to be something obvious...lol
$start = "2014-05-22 09:16:24";
$type = 30;
$endtime = date($start,strtotime("+ '.$type.' minutes"));
echo "End: ".$endtime."</br>Start: ".$start;
Right now it returns:
End: 2014-05-22 09:16:24
Start: 2014-05-22 09:16:24
when i use strtotime('+3 HOURS') everything is fine. so why cant i use strtotime('+3.5 HOURS')? what would be the proper way to do this? 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! how do I use strtotime() instead of mktime()? $matchTimea = mktime($mHour, $mMinute, 0, $mMonth, $mDay, $mYear); if($matchTimea-time() < $cutoffTimea*60) { 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? 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 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? 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. 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 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 hello, what am i doing wrong? Code: [Select] $starttime=$row['starttime']; $esttime1=date( 'g:i a', strtotime(-2 hours, $starttime) ); 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 ! |
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