PHP - Days Left To Include Leap Year.

Is there a recommended way to code in order to have a dual language support, especially when one is English (left to right), and the other is Hebrew (right to left)?

Should I design every page twice or is there a ready made solution?

Why does this work:
$dateAllowedSelect = date('Y') -18;

But this doesn't:
$dateAllowedSelect = date('Y') +5;

I nee to get the current year plus 2010 - 2015 in a select box.

Thank you.

I am using the following code to sort by Year. It displays the current year fine and current year also displays in selection but does give the option for the previous years selection.

Code: [Select]
<form action="archivednews.php" method="post">
<select name="year" id="year">
<?PHP for($i=date("Y");
$i<=date("Y");
$i++)  
if($year == $i)
echo "<option value='$i' selected>$i</option>";
else
echo "<option value='$i'>$i</option>";
?>
</select>
<input type="submit" value="GO">
</form>

Gud Pm!
Im trying to display a list of years (last year, current year and next year) with these simple code i made..
It displays the current and next year correctly except last year, which gives me "1970".
I don't get why it gives me that since may next year works properly.. 

Code: [Select]
$currentyear = date('Y');
$subtract_year = strtotime(date('Y', strtotime($currentyear)) . '-1 year');
$lastyear = date('Y', $subtract_year);
$addyear = strtotime(date('Y', strtotime($currentyear)) . '+1 year');
$newyear = date('Y', $addyear);
$year_array = array($lastyear, $currentyear, $newyear);
foreach($year_array as $year)
{
echo "<option value='{$year}'>";
echo $year;
echo "</option>";
}

I'm not familiar with dates yet.. so i'm gonna read the manual while w8ting for your kind replies. 

I need help with a query.   I've got a table that stores references to calls that come in to our office. It has a datetime field, and for the purposes of the query, it's the only thing in this table that is important.   I've got another table that stores appointment information. The appointment start time is also a datetime field. For this query, this start time is the only thing in this table that is important.   I need to know how many times in the last year there was a day when we had at least 150 calls come in and made at least 1 appointment.

SELECT COUNT(*)
FROM appointments a
LEFT JOIN (
  SELECT DATE(time)
  FROM calls
  WHERE <there are at least 150 calls on a day>
) c 
ON DATE(a.start_time) = c.time
WHERE a.start_time >= CURDATE() - INTERVAL 1 YEAR 

To make things difficult, I guess the time would have to be converted to a day, because I'm looking for all days where there was an appointment, and the datetime field is too specific.   I've been looking online for 30 mins or so, and was hoping somebody here could point me in the right direction. I'm not making much progress by myself.
Edited by sKunKbad, 06 October 2014 - 01:17 PM.

Okay I am trying to list out years in a dropdown box. It should show say how many years old. I want to start from 16 though and stop at 100. So the year has to be 1995 for 16 but then I would have to change the code every year. so I was wondering how do I modify this code to do what I explained?
Code: [Select]
<?php
 $start_year = ($start_year) ? $start_year - 1 : date('Y') - 100;
    $end_year = ($end_year) ? $end_year : date('Y');
for ($i = $end_year; $i > $start_year; $i -= 1) {

$date=date(Y);
$age = $i - $date;

        echo '<option value="'.$i.'">'.$age.'</option>';
    }
?>

Thanks in advanced

So I have decided to do research as my project and gain something from this. But also I will have to include programming which i prefer PHP. and will be testing these for my research. Does this sound like a good project and also If it is good sound.. Where do I start?...  need a starting point and then I could go on to doing it..

Hello,

I seem to be having a problem.  I am trying to extract the year from a date

Code: [Select]
2012-03-01
echo "2012";

I have tried this and it only displays 1969

$dateorig = "2012-03-01";
$new_year = date("Y", strtotime($dateorig));
echo $new_year;

I'm trying to figure out if today's current date and time falls between Mon & Fri, but so far, I can't get it to work.
Can anybody look at my code and see what I'm doing wrong?

Code: [Select]
<?php
$day_start = date('D h:i a', strtotime("Mon 05:30"));
$day_end = date('D h:i a', strtotime("Fri 10:00"));
$day_current = date('D h:i a', strtotime("+1 hours"));

if (($day_current > $day_start) && ($day_current < $day_end))
{
echo "yup";
}
else {
echo "nope";
}
?>

Thanks in advance

I'm using this code to get me the variables to a select box on a register page:
$vars['user-birthdate-years'] = array(_VARS_USER_CHOOSE_ => "Please choose", _VARS_USER_NONE_ => _VARS_USER_UNDEF_); for ($i = date('Y') - 18; $i > date('Y') - 75; --$i) { $vars['user-birthdate-years'][$i] = "$i"; }

This gives me this result in the page source:
<select name="birthyear" class="date"><option value="-2" selected="selected">Please choose</option><option value="1997">1997</option><option value="1996">1996</option><option value="1995">1995</option><option value="1994">1994</option><option value="1993">1993</option><option value="1992">1992</option><option value="1991">1991</option><option value="1990">1990</option><option value="1989">1989</option><option value="1988">1988</option><option value="1987">1987</option><option value="1986">1986</option><option value="1985">1985</option><option value="1982">1982</option><option value="1981">1981</option><option value="1980">1980</option><option value="1979">1979</option><option value="1978">1978</option><option value="1977">1977</option><option value="1976">1976</option><option value="1975">1975</option><option value="1974">1974</option><option value="1973">1973</option><option value="1972">1972</option><option value="1971">1971</option><option value="1970">1970</option><option value="1969">1969</option><option value="1968">1968</option><option value="1967">1967</option><option value="1966">1966</option><option value="1965">1965</option><option value="1964">1964</option><option value="1963">1963</option><option value="1962">1962</option><option value="1961">1961</option><option value="1960">1960</option><option value="1959">1959</option><option value="1958">1958</option><option value="1957">1957</option><option value="1956">1956</option><option value="1955">1955</option><option value="1954">1954</option><option value="1953">1953</option><option value="1952">1952</option><option value="1951">1951</option><option value="1950">1950</option><option value="1949">1949</option><option value="1948">1948</option><option value="1947">1947</option><option value="1946">1946</option><option value="1945">1945</option><option value="1944">1944</option><option value="1943">1943</option><option value="1942">1942</option><option value="1941">1941</option></select>

Problem: When running the php code at the top it miss two years in the span as you see in the page source below, the year 1983 and 1984 is completely missing in the page source. Why is this happening, could the php code be corrected or is there a workaround in php for the missing years bug I seem to have found?

I really appreciate any help on this!
Edited by Drezzzer, 12 January 2015 - 05:16 AM.

Hello there,

I stuck in a silly logic ... I will really appreciate if someone helps me...

My Fiscal year starts from 01st April of current year through 31st March of next year, for ex. (01 Apr 2011 to 31st March 2012). I want to put a latefee of 6.25% on total amount, if the payment date is not done on the current fiscal year. Everything depends on the payment date.

For example,

A bill of $8000 was generated on 01st Apr. 2009 then its 6.25% will be $500 and the customer is paying the same on 30th August 2011, then total years passed is 3. So, the late fee will be 500 * 2 = $1000.

i.e. Any bill generated and paid in the current fiscal year (stated above) is free of late fee and after that the above calculation is applicable.

Please help, how to do the calculation. A function is really appreciated.

Thanks to aleX_hill and RussellReal for the help so far. I'm trying to utilise the code before to find the next month. However, it needs to be used with the current year. So I altered the below


<?php
$nextMonth = date("m") + 1;
$daysInNextMonth = date("t",mktime(0,0,0,$nextMonth));
echo $daysInNextMonth;?>


to

#NEXT MONTH
$nextMonth = date("m") + 1;
echo $nextMonth; 
$daysInNextMonth = date("t",mktime(0,0,0,$nextMonth));
echo $daysInNextMonth ;
$year = date("o"); 
echo $year;


I can then pass it into my url query like so:

<a href="<?php echo $url_path; ?>after=<?php echo $year; ?>-<?php echo $nextMonth;  ?>-01&before=<?php echo $year; ?>-<?php echo $nextMonth;  ?>-<?php echo $daysInNextMonth; ?>">Next Month</a>


Brilliant I thought, until it comes to December 2010, the next month needs to be January 2011. Now I'm bamboozled. How do I get around this one?

Many thanks, your help is always appreciated!

I have a given date on a webpage, that I'm scraping to insert into a DB.  The date is in this format:

Sun, Feb 9<br />3:00 PM ET

I need to insert this into the DB in this format:  2020-02-09 15:00:00

It will always be this year...How can I change this data to be inserted correctly?

I'm trying this...and it's inserting as 1969

	$date = 'Sun, Feb 9<br />3:00 PM ET':
$healthy = array("<br />", " ET", "Sun.", "Mon.", "Tue.", "Wed.", "Thu", "Fri.", "Sat.", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec");
$yummy   = array("  ", "2020", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December");
	$date = str_replace($healthy, $yummy, $date1);
	$start = date('Y-m-d H:i:s', strtotime("$date -120 minutes"));
	

Thanks!

Hello this code is supposed to sort by year and month, but it doesn't.  It sort by year and months are random.  Any suggestions?  Many thanks,

$sortdata =[];
        $namecolumn = array_column($data, '0');
        foreach($data as $datecolumn){
            $dateexplode = explode('/', $datecolumn[3]);
            $sortdata[] = (isset($dateexplode[1])) ? $dateexplode[1] : ((isset($dateexplode[0])) ? $dateexplode[0] : '');
    
        }
        array_multisort($namecolumn, (($sorttype=='DESC') ? SORT_DESC : SORT_ASC), $sortdata, (($sorttype=='DESC') ? SORT_DESC : SORT_ASC), $data);

            

Edited May 13 by Barand
code tags added

When I try to add 30 days:
Code: [Select]
$date = date("Y-m-d");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days");
echo $date;
and I echo date I get 1330664400

How do I get it to echo out 3/1/2012?

I know the answer lies in the strtotime but I can't figure it out.  I know it's a simple problem for most of you...