|
PHP - Is There A Way To Remove The + Sign From Date Diff? And Is It Possible To Convert It To Months?
So this is a simple code that finds out the difference between two dates and displays it in number of days.
$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");
// RESULT
+272 days
My first question. Is it possible to remove the + sign in the result above? Second question. Is it possible to show "months" if it's greater than 30 days? And years if the days are greater than 365? How would I do this? Similar TutorialsHi Everyone I have an older coded side that I am looking at for a friend, I need to add in a date diff colum into their existing index page. I can add using the following code onto the indivudually record page can't get it to work on the index. Any help is appreciated. Code on Individual page which works:::
<HEAD>
<?
if($test['settlement_date']=="0000-00-00"){
$date1 = new DateTime($date_received);
$date2 = new DateTime("now");
$interval = $date1->diff($date2);
} else {
$date1 = new DateTime($date_received);
$date2 = new DateTime($settlement_date);
$interval = $date1->diff($date2);
}
?>
</HEAD>
<Body>
<?php echo "".$interval->days."";?>
</BODY>
I was trying to add something similar but as it is a list I can't work out what I need to do. INDEX:::
while($test = mysql_fetch_array($result))
{
$complaint_id = $test['complaint_id'];
echo"<tr>";
echo"<td class='standard_left'>".$test['complaint_id']."</td>";
echo"<td class='standard_left'>".$test['status']."</td>";
echo"<td class='standard_left'>".$test['LastName'].", ".$test['FirstName']."</td>";
echo"<td class='standard_left'><a href ='complaint_view.php?complaint_id=$complaint_id'>".$test['complainant']."</td>";
echo"<td class='standard_center'>".$test['date_received']."</td>";
echo"<td class='standard_center'>".$test['settlement_date']."</td>";
echo"<td class='standard_center'>".$test['interval->days']."</td>";
echo "</tr>";
I know it is old redundent code but if I can get this one colum working I would be happy. I am working on a report generation. Here I need to count the number of months involved in the selected date range. I need to apply the monthly charges accordingly. Example: If the user selects 25-08-2011 to 03-10-2011, it should return 3 as the number of months. Yes, the number of days are less than 60 but still the date range is spread across 3 months. Any solution.. Please. Girish 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 using the following code to get the third sunday of every month for +6 Months from todays date. I have put together the following code which works perfectly when the date is echoed however when it comes to inserting those dates into mysql db it throws a unable to convert to string error. Im new to this and my head is in a spin as to why those dates can't be inserted if they can echoed ok. I have spent 2 days at it now and searched many forum however they seem to get side tracked from my issue (or do they)??????? I figure I can't go mixing DateTime and date() the way I have? Please help??? Code: [Select] [php] $month=date('F'); $now=date('Y-m-d'); $year=date('Y'); $num='3'; //example only this will be a $_POST value 0-3 $day='Sunday'; //example only this will be a $_POST value Monday-Sunday $start= date('Y-m-d', strtotime('+'.$num.' week '.$day.' '.$month.' '.$year.'')); if($start>$now) { $begin=strtotime('+'.$num.' week '.$day.' '.$month.' '.$year.''); } else { $d = new DateTime( $start ); $d->modify( 'first day of next month' ); $nextmonth = $d->format( 'F' ); $year = $d->format( 'Y' ); $begin=strtotime('+'.$num.' week '.$day.' '.$nextmonth.' '.$year.''); } $date=date('Y-m-d', $begin); # insert into db first occurance //mysql_select_db($database); //$query_rsFirst = "INSERT INTO event_date VALUES (NULL, '$event', '$d', '$date', '$date')"; //$rsFirst = mysql_query($query_rsFirst) or die(mysql_error()); echo $date.'<br/>'; for($i = 0; $i <= 4; $i++) { $d = new DateTime( $date ); $d->modify( 'first day of next month' ); $nextmonth = $d->format( 'F' ); $date=date('Y-m-d', strtotime('+'.$num.' week '.$day.' '.$nextmonth.' '.$year.'')); # insert into db //mysql_select_db($database); //$query_rsDate = "INSERT INTO event_date VALUES (NULL, '$event', '$d', '$date', '$date')"; //$rsDate = mysql_query($query_rsDate) or die(mysql_error()); echo $date.'<br/>'; if($nextmonth=="December") { $d = new DateTime( $date ); $d->modify( 'first day of next year' ); $year = $d->format( 'Y' ); } } [/php] $username = $_POST['username']; $password = $_POST['password']; $month = $_POST['month']; $day = $_POST['day']; $year = $_POST['year']; $query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$month','$day','$year') mysql_query($query); The code above is a sample of what I have but what I want is to store an entire birthdate in ONE SQL cell. More like this... $username = $_POST['username']; $password = $_POST['password']; $month = $_POST['month']; $day = $_POST['day']; $year = $_POST['year']; $query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$birthdate') mysql_query($query); How is this possible? Can I do this and actually use it efficiently in the future? I have tried a large number of "solutions" to this but everytime I use them I see 0000-00-00 in my date field instead of the date even though I echoed and can see that the date looks correct. Here's where I'm at: I have a drop down for the month (1-12) and date fields (1-31) as well as a text input field for the year. Using the POST array, I have combined them into the xxxx-xx-xx format that I am using in my field as a date field in mysql. <code> $date_value =$_POST['year'].'-'.$_POST['month'].'-'.$_POST['day']; echo $date_value; </code> This outputs 2012-5-7 in my test echo but 0000-00-00 in the database. I have tried unsuccessfully to use in a numberof suggested versions of: strtotime() mktime Any help would be extremely appreciated. I am aware that I need to validate this data and insure that it is a valid date. That I'm okay with. I would like some help on getting it into the database. 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 In sql we have dates like this 134238 In php it shows - 19:10:11 . we can make it by using date('d.m.y', $date); where $date=134238 ; my question is how can we turn 19:10:11 in to 134238 using PHP Is it possible to convert time(); to date();? while I'm reading the cell value '20/09/1980' from an excel file using phpexcel class method $student[$i]['d_o_b'] = $objWorksheet->getCell('G' . $intRow)->getValue(); getting the value as '29484' pls help me to convert it into the date format again. Hi all, I'm having a bit of trouble a script running on a site where it converts a date of birth in a database shown like this '30/04/1993' to an actual age, for instance 18 in this case. Only the script I'm using below shows this age as 17, not 18 as it should be. Code: [Select] <?php $birthday = $row_getdets['dob']; function birthday ($birthday){ list($day,$month,$year) = explode("/",$birthday); $day_diff = date("d") - $day; $month_diff = date("m") - $month; $year_diff = date("Y") - $year; if ($day_diff < 0 || $month_diff < 0) $year_diff--; return $year_diff; } ?> So i've tried to remedy this myself with the following: Code: [Select] <?php $birthday = $row_getdets['dob']; function birthday ($birthday){ list($year,$month,$day) = explode("/",$birthday); $year_diff = date("Y") - $year; $month_diff = date("m") - $month; $day_diff = date("d") - $day; if ($month_diff < 0) $year_diff--; else if (($month_diff==0) && ($day_diff < 0)) $year_diff--; return $year_diff; } ?> ..but I'm having a syntax error (unexpected T_LINE), most probably down to my novice ability, I bet I've missed something simple. I'm still learning guys and I'd really appreciate any help at all. 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 sorry, I know this is simple, but stuck on it... how would one go about converting 1/21/11 to a UNIX timestamp, something like 1293035229? and while we are at it, how to convert 1293035229 to 1/21/11 ? (i know the timestamp is wrong, just for example) thanks!!!!! 2011-02-05T18:07:57.00+0000 This is a string. hello, i have the following code and it is ok. what im trying to do is convert a 2010-09-20 post from a form and have it read out as Monday September 20, 2010. is this possible? Code: [Select] $dateselected="$_POST[Y]-$_POST[M]-$_POST[D]"; echo "<table border='1'>"; echo "<tr><td width='100%' colspan='4' align='center'>$_POST[M]-$_POST[D]-$_POST[Y]</td></tr>"; Hello, i need a fast convertor that converts the date format 03-03-2008, 08:26 PM into a timestamp fast. Like, you paste 03-03-2008, 08:26 PM into a post form, submit and it gives you the time stamp. I made a convertor a whille ago, but you need to insert each numbers into a separate input box (day, months, year etc) I cant remember how exactly it functions etc asi haventtouched PHP In a while..... Hi. I have an input field where a date is entered, format dd-mm-yy. I need to query the database to see if this date exists. How can I convert the date to yyyymmdd before the query? Thanks My query is a fixtures list for my local sports team- There seems little point including fixtures from the past as they are in a results query anyway. I'm ordering by date (see below) but how can I remove the ones already past? $query = "SELECT * FROM fix10 ORDER BY Date"; TIA Nick Hey all, Im currently working on a site which I save a cookie on users computer for 1 month as well a row in a table with the 1 month expiring date plus cookie id. What I would like to know is if theres a way to make the server check the table every day at say 12pm to see if any corresponding user cookies in each row have expired on that day as to remove them from the table to match the expiring cookie on the users computer? I would like not to have to rely on a user/admin accessing in to make a script run instead have something thats timed to go off automatically instead? Is there anyway of doing this? Thanks! I wish to know how i can convert this date 24-09-2010 to this format 2010-09-24 |
|||||||