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PHP - Taking An Hour Off A Date String
Hi,
I am creating a date string using the following: $curDate = date("Y-m-d H:i:s"); However, what I want to do is take one hour off this so that $curDate now becomes the current date and time minus one hour. Thanks for your help. Similar TutorialsHello, Im trying to work out some code. On my site between Thursday 12th July and Sunday 15th July between the hours of 22:00 and 23:00 I want to display some code. I've done this so far, but am having trouble. Anyone please help? $the_date = date("H:i:s"); $timesep = explode(":",$the_date); // $hour = $timesep[0]; if (($the_date > 2010-08-12 && < 2010-08-15) && ($hour >= 22) && ($hour <= 23)) { //Code } Hi, I am trying to convert a String date into numeric date using PHP function's, but haven't found such function. Had a look at date(), strtotime(), getdate(); e.g. Apr 1 2011 -> 04-01-2011 Could someone please shed some light on this? Regards, Abhishek Hi Guys, I have got a date string which looks like the following: Quote Thu Feb 09 2012 07:25:00 GMT 0000 (GMT Standard Time) I need to convert it to Quote d-m-Y H:i format. Any ideas on how to do this? Thanks! PLEASEEE no matter what i do... datepicker always post the dates as string i do whatever ways know convert string to date but no way!! how convert string to date so i can sarch in a mysql table... tnx in advanced <!--HTML-->
<div class="control-group"> <!--PHP-->
$yfromx=$_POST['from'] ;
$t1=gettype ( $yfromx ) ;
I have another question related to string searches. What I have is a string of couple thousand characters of all types but the date format is allways the same: dd-mth-year (two digits for day-three letters for month-four digits for year). Can I use the format alone to determine where the date is located which in turn would make it very easy to store that date into variables for inserting into database, etc. Is there anyway to tell php the format to look for or do I have to have actual numbers/letters to perform the search? Hi, Sorry, me again! Ok, so i've got my url: http://www.mydomain.com/?ec3_after=2010-08-01&ec3_before=2010-08-07 I've then got the following code to get the values: $afterDateParts = split("-", $_GET['ec3_after']); $after = $afterDateParts[2] . " " . $afterDateParts[1] . " " . $afterDateParts[0]; echo $after; which returns: 01 08 2010 (correct for this demo) I've then got this to convert the above into a nicer format: $convertMe = strtotime($after); echo date('d-M-Y', $convertMe); But it always returns: 01 Jan 1970 For the love of god, I cannot work out why. Is it really not that simple? Please someone put me out my misery. I'm loosing hair by the minute! Lol TIA I'm getting this error "Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/westiehi/public_html/groupBOS.php on line 36" for this code: Code: [Select] $query_rs_specialty = "SELECT * FROM specialty WHERE Group_Place <> '' and Date>= DATE_SUB(CURDATE(), INTERVAL 6 MONTH) ORDER BY Date DESC"; I've tried everything to get this to take records where the one field is blank and the date is greater than a date 6 months ago but nothing seems to work. Help would sure be appreciated!! I am trying to display date and time below code display date correctly but when it come to display time it display 3 hour behind. eg. when it is 12:47 PM in my system it display 9:47 PM Code: [Select] <?php $todaydate=date("j - F - Y g:i:s a"); echo $todaydate; ?> $time="19:00:00"; how to do i output one hour later from the variable above? Hi, i'm trying to create some detailed statistics about customer activity, i have entries in my mysql db when the customer has been active for the last time and want to create some statistics about that, basically a "online last 24 hours" but from specific countries. Now i've tried this: Code: [Select] $time = date('Y-m-d H:i:s'); $time24 = date("Y-m-d H:i:s", time()-((60*60)*24)); $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de' AND time BETWEEN '$time' AND '$time24' "; to get the current date and select all customers that have been available from the current date minus 24 hours, what's my mistake here, as this doesn't seem to work! Thanks Hi guys, I have a PHP Script were user would enter a username and password to get access, but the problem I keep having is every half n hour or so user have to sign in again cause the keeps logging them out for some reason. Please I need your help and advice to make this stop. Hi I want to make something like this - My Sites index.php will be avail avail to user after he has clicked in a link that will come after every 24 Hour in my site. Means when a user first enters the site it will come and clicking in there the site will be avail avail. again after 24 Hour it will come again. But i am not getting how to do it. So need help SaKIB Hello, I am trying to load an external image from a NOAA site directory. This image refreshes every three hours, starting with 0 and then 3,6,9,12.....21 in a 24 hour calendar, and are stored for days in advance. The date/time is in the file name. Originally I had written the script to refresh on the hour, and then realized the tri-hourly spacing. This is what I have: Code: [Select] <?php date_default_timezone_set('Etc/GMT-1'); $today = date("YmdH"); $pattern = '/WaveHeight_'.$today.'_mic.png'; $base_url = 'http://www.crh.noaa.gov/images/greatlakes/ndfd/MIC/dynamic2'; print '<a href="http://www.crh.noaa.gov/greatlakes/?c=map&l=lm&p=a"><img src="'.$base_url.$pattern.'" width= "237" hieght= "314" " ></a>'; ?> I am not fluent in php, but I assume I need a "If H=1,4,7,10,13,16,19,22; Then H+2 ; along with If H=2,5,8,11,14,17,20,23; Then H+1" to keep the images relatively current. I just can't quite wrap my head around how to express this. Any help is appreciated. Thank you. Hi,
I have a blog which records the amount of views on each article. I now want to be able to work out the average number of views per hour.
How can I work out the number of hours passed from a datetime format?
From there I can just do views divided by hours passed.
Thanks in advance!
Hello, I have a very simple table... DAY (Y-m-d) / TIME (00:00:00) / POWER (INT) I am using a Jquery datepicker to select the date on the page, and that POST to the PHP file. I am trying to select values from Mysql to make 3 HighCharts Graphs using the selected day of the datepicker. I have started with the DAY graph PHP to get the values for each hour of a 24 hour day. I need to get the values for each hour... 23,12,15,35 etc... , and then I need for the 31 days of a month for the month graph, and 12 months of the year for the month graph all in the same way so the HighCharts can use the values to make the chart (3 php files, one for each graph) Here is the PHP I have for the 1 day that should get the 24 individual hour data, but it does not seem to work... <?php $choice = (isset($_POST['choice'])) ? date("Y-m-d",strtotime($_POST['choice'])) : date("Y-m-d"); $con = mysql_connect("localhost","root","mackie1604"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("inverters", $con); $sql = "SELECT HOUR(time), COUNT(power) FROM feed WHERE time = '".$choice."' GROUP BY HOUR(time) ORDER BY HOUR(time) "; $res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error()); $row = mysql_fetch_assoc($res); echo $row['choice'].'<br />'; ?> What have a written wrong in the sql query ??? Alan Hi all, I have a loop that is modifying a DateTime object. It goes through X number of times and adds 1 day to the date time object and stores the modified object in an array. I use ->add(new DateInterval('P1D')); For some reason it is losing an hour each time that I add a day. I start out at 19:00 and each iteration through the loop adds a day, but loses an hour. 2010-09-19 19:00:00 2010-09-20 18:00:00 2010-09-21 17:00:00 I can correct for this by doing ->add(new DateInterval('P1DT1H'));, but adding 25 hours instead of 24 hours doesn't make any sense to me. I am using PHP Version 5.3.1. Any ideas? Thanks, W So I am trying to check if the current time is one hour before a variable time: Code: [Select] $date_game=$dt->format('Y n j'.$pieces[2]); echo $date_game; echo date('Y n j H'); if (date('Y n j H') < $date_game) { echo "The time is before the stored time"; } This displays 2012 1 21 17:30 2012 1 21 16 i.e $pieces[2] = 17:30. and $dt formatted Y n j = 2012 1 21. The current Y n j H is 2012 1 21 16. I want to know if it is more than one hour until the date/time stored in $date_game. At the moment it just tells me that it is before that time. Can I do something like Code: [Select] $data_game - 1->format('H'); or something? Thanks guys Hi guys I was just wondering if there is some date combination for which this type of comparisons will not work. var_dump("2010-15-1">"2010-4-01"); Provided the date is always in the YYYY-mm-dd format with optional preceding 0 . Thanks This topic has been moved to PHP Freelancing. http://www.phpfreaks.com/forums/index.php?topic=332844.0 I insert a time value using the following code
$ti= ( !empty($_POST['time']) ) ? "'{$_POST['time']}'" : 'NULL';
I want to insert a second time value into a differen column which would be the same time minus 1 hour
something along the lines of :
$ti2= ( !empty($_POST['time']) ) ? "'{$_POST['time'] -1 hour}'" : 'NULL';
What would be the correct way to do it
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