I want to only shows result for the current month, but I have no idea how to do this, my current output is like this.
Output
2014-04-11
-> array
2014-04-11
2014-04-05
2014-03-29
PHP
$date = date('Y-m-d');
echo $date;
echo "<pre>";
foreach ($submissions as $test){
if($date >= substr($test['thing']['created'], 0, 10)){
echo substr($test['thing']['created'], 0, 10);
echo "<br>";
}
}
echo "</pre>";
My current code wont work as its only checking if the whole number is greater or equal to, any ideas anyone?
Just try with strtotime:
$year = date('Y');
$month = date('m');
foreach ($submissions as $test) {
$timestamp = strtotime($test['thing']['created']);
$testYear = date('Y', $timestamp);
$testMonth = date('m', $timestamp);
if (($month >= $testMonth && $year == $testYear) || $year > $testYear) {
// test passed
}
}
If your problem is to have only results of current month :
$date = date('Y-m');
echo "<pre>";
foreach ($submissions as $test){
// if month & year is equal
if($date == substr($test['thing']['created'], 0, 7)){
echo substr($test['thing']['created'], 0, 10);
echo "<br>";
}
}
echo "</pre>";
If you want to sort it, you will have to convert to timestamp
i'm not a pro in php but take a look here : http://fr2.php.net/manual/fr/function.date.php
does date('m') would be a solution for your problem?
You can try 2 ways :
convert them to timestamp, then compare the int
If you are taking the dates frome a DB, ORDER BY date in the query
Variable for current month first $cur_m=date('m')
Explode your variable in loop with -.
$p = explode('-',$test['thing']['created']);
in loop check for current month like below
if($p[1]==$cur_m){ your code... }
Related
I have a php code as shown below in which on the 1st day of every month, I am copying 2nd JSON object array (next_month) content into 1st JSON object array (current_month).
In the 2nd JSON object array (next_month), I want to have next month dates. That will also happen on the 1st day of every month. Currently I am storing nada. Let us suppose that today is 1st day of November.
php code:
$value = json_decode(file_get_contents('../hyt/dates.json'));
if ((date('j') == 1)) {
$month = 11;
$year = date('Y');
$current_month_days = (date('t', strtotime($year . '-' . $month . '-01')));
$next_month_days = (date('t', strtotime($year . '-' . ($month + 1) . '-01')));
$value->current_month = $value->next_month; // Line Y
$value->next_month = array_fill(0, ($next_month_days), nada); // Line Z
}
The current look of JSON (dates.json) is shown below:
{"current_month": ["2020-10-01", "2020-10-02", "2020-10-03", "2020-10-04", "2020-10-05", "2020-10-06", "2020-10-07", "2020-10-08", "2020-10-09", "2020-10-10", "2020-10-10", "2020-10-12", "2020-10-13", "2020-10-14", "2020-10-15", "2020-10-16", "2020-10-17", "2020-10-18", "2020-10-19", "2020-10-20", "2020-10-21", "2020-10-22", "2020-10-23", "2020-10-24", "2020-10-25", "2020-10-26", "2020-10-27", "2020-10-28", "2020-10-29", "2020-10-30","2020-10-31"],
"next_month": ["2020-11-01", "2020-11-02", "2020-11-03", "2020-11-04", "2020-11-05", "2020-11-06", "2020-11-07", "2020-11-08", "2020-11-09", "2020-11-11", "2020-11-11", "2020-11-12", "2020-11-13", "2020-11-14", "2020-11-15", "2020-11-16", "2020-11-17", "2020-11-18", "2020-11-19", "2020-11-20", "2020-11-21", "2020-11-22", "2020-11-23", "2020-11-24", "2020-11-25", "2020-11-26", "2020-11-27", "2020-11-28", "2020-11-29", "2020-11-30"] }
Problem Statement:
I am wondering what changes I should make at Line Z so that in the second JSON object array, I am able to get next month dates. At present, I am storing nada.
The content which I want in the JSON on the 1st day of November month after successful execution of Line Y and Line Z is:
{"current_month": ["2020-11-01", "2020-11-02", "2020-11-03", "2020-11-04", "2020-11-05", "2020-11-06", "2020-11-07", "2020-11-08", "2020-11-09", "2020-11-11", "2020-11-11", "2020-11-12", "2020-11-13", "2020-11-14", "2020-11-15", "2020-11-16", "2020-11-17", "2020-11-18", "2020-11-19", "2020-11-20", "2020-11-21", "2020-11-22", "2020-11-23", "2020-11-24", "2020-11-25", "2020-11-26", "2020-11-27", "2020-11-28", "2020-11-29", "2020-11-30"],
"next_month": ["2020-12-01", "2020-12-02", "2020-12-03", "2020-12-04", "2020-12-05", "2020-12-06", "2020-12-07", "2020-12-08", "2020-12-09", "2020-12-11", "2020-12-11", "2020-12-12", "2020-12-13", "2020-12-14", "2020-12-15", "2020-12-16", "2020-12-17", "2020-12-18", "2020-12-19", "2020-12-20", "2020-12-21", "2020-12-22", "2020-12-23", "2020-12-24", "2020-12-25", "2020-12-26", "2020-12-27", "2020-12-28", "2020-12-29", "2020-12-30", "2020-12-31"] }
This is what I have tried:
This is what I have tried at Line Z but its storing only today's date in JSON object array.
$value->next_month = array_fill(0, ($next_month_days), date("Y-m-d")); // Line Z
I think you should completely recreate your JSON string. It starts on the first day of the current month. The loop always runs as long as the month remains. The whole thing then again for the following month.
$arr = $cur = [];
$date = date_create('first day of this month 00:00');
$startMonth = $month = $date->format('m');
while($startMonth == $month){
$cur[] = $date->format('Y-m-d');
$date->modify('+1 Day');
$month = $date->format('m');
}
$arr["current_month"] = $cur;
$startMonth = $month;
$cur = [];
while($startMonth == $month){
$cur[] = $date->format('Y-m-d');
$date->modify('+1 Day');
$month = $date->format('m');
}
$arr["next_month"] = $cur;
$jsonStr = json_encode($arr);
You are using array_fill, which is used to fill at least part of an array with the same value. I would recommend using a simple for loop:
$next_month_array = [];
$next_month = $month < 12 ? $month + 1 : 1;
$year = date('Y');
for($day_counter = 1; $day_counter <= $next_month_days; $day_counter++) {
$next_month_array[] = "$year-$next_month-$day_counter";
}
$value->next_month = $next_month_array;
I am trying to show results of each month.
Im having this for loop:
foreach ($overview as $day) {
$year = date("Y") - 1;
if ($day->user == $info->id) {
$startDate = new DateTime($day->Calendar_startdate);
$endDate = new DateTime($day->Calendar_enddate);
$s = $startDate->format('Y-m-d');
$e = $endDate->format('Y-m-d');
if ($s > $year) {
$workdays = number_of_working_days($s, $e);
$daysleft = $daysleft + $workdays;
} else {
}
}
}
This for loop is also in an if statement which echos the months.
Now I need to let it work for the months January, February etc...
I am able to not show results if in the previous year which works well.
If you want to compare $s with $year just change $year to :
$time = new DateTime('now');
/*** you can use `now` for today
/* or you can change to a fixed date exmp: 2016-01-01
*/
$year = $time->modify('-1 year')->format('Y-m-d');
Than you can compare $s > $year
I fixed by checking each month if it contained for example -01-
DB::table('Calendar')->where('Calendar_startdate', 'like','%' . $monthnumber . '%')->where('user', $info->id)->where('Calendar_type',2)->get();
I have a PHP script which records things based on the day. So it will have a weekly set of inputs you would enter.
I get the data correctly, but when i do $day ++; it will increment the day, going passed the end of the month without ticking the month.
example:
//12/29
//12/30
//12/31
//12/32
//12/33
Where it should look like
//12/29
//12/30
//12/31
//01/01
//01/02
My script is as follows:
$week = date ("Y-m-d", strtotime("last sunday"));
$day = $week;
$run = array(7); //this is actually defined in the data posted to the script, which is pretty much just getting the value of the array index for the query string.
foreach( $run as $key=>$value)
{
$num = $key + 1;
$items[] = "($num, $user, $value, 'run', '$day')";
echo "".$day;
$day ++;
}
Should I be manipulating the datetime differently for day incrementations?
You can use
$day = date("Y-m-d", strtotime($day . " +1 day"));
instead of
$day++;
See live demo in ideone
You refer to $day as a "datetime" but it is just a string - that is what date() returns. So when you do $day++ you are adding 1 to "2015-12-02". PHP will do everything it can to make "2015-12-02" into a number and then add 1 to it, which is not date math. Here is a simple example:
<?php
$name = "Fallenreaper1";
$name++;
echo $name
?>
This will output:
Fallenreaper2
This is how I would do it, using an appropriate data type (DateTime):
<?php
$day = new DateTime('last sunday');
$run = array(7);
foreach ($run as $key => $value) {
$num = $key + 1;
$dayStr = $day->format('Y-m-d');
$items[] = "($num, $user, $value, 'run', '$dayStr')";
echo $dayStr;
$day->modify('+1 day');
}
To increase time you should use strtotime("+1 day");
here is simple example of using it
<?php
$now_time = time();
for($i=1;$i<8;$i++) {
$now_time = strtotime("+1 day", $now_time);
echo date("Y-m-d", $now_time) . "<br>";
}
?>
I only want to show some html on certain days of the week. I think I am very close, but cannot get this to work. Thanks for the help.
<?php
$dayofweek = date('l');
$daystoshow = array('Thursday','Sunday','Wednesday');
if ($dayofweek == $daystoshow) {
echo "show this";
}
?>
Use in_array() to check to see if a value in in an array:
if (in_array($dayofweek, $daystoshow)) {
Allright:
Lets use PHP date function
Do you want to show it only during a particular month ?
$date = date("m");
if ($date == "01") { // only on january
//output
}
Do you want to show it only during a particular day ?
$date = date("d");
if ($date == "01") { //Only on the first of the month
//output
}
Do you want to show it only during particular dayS ?
$days = array("01","11","28");
$date = date("d");
if (in_array($date,$days)) { //Only 01,11,28
//output
}
Do you want to show it only during particular days of week ?
$days = array("Thursday","Sunday","Wednesday");;
$date = date("l");
if (in_array($date,$days)) {
//output
}
Do you want to show it only during a particular year ?
$date = date("Y");
if ($date == "2014") { //Only on 2014
//output
}
I want to take a date and work out its week number.
So far, I have the following. It is returning 24 when it should be 42.
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[2], $duedt[1],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
Is it wrong and a coincidence that the digits are reversed? Or am I nearly there?
Today, using PHP's DateTime objects is better:
<?php
$ddate = "2012-10-18";
$date = new DateTime($ddate);
$week = $date->format("W");
echo "Weeknummer: $week";
It's because in mktime(), it goes like this:
mktime(hour, minute, second, month, day, year);
Hence, your order is wrong.
<?php
$ddate = "2012-10-18";
$duedt = explode("-", $ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2], $duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: " . $week;
?>
$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
Use PHP's date function
http://php.net/manual/en/function.date.php
date("W", $yourdate)
This get today date then tell the week number for the week
<?php
$date=date("W");
echo $date." Week Number";
?>
Just as a suggestion:
<?php echo date("W", strtotime("2012-10-18")); ?>
Might be a little simpler than all that lot.
Other things you could do:
<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>
Becomes more difficult when you need year and week.
Try to find out which week is 01.01.2017.
(It is the 52nd week of 2016, which is from Mon 26.12.2016 - Sun 01.01.2017).
After a longer search I found
strftime('%G-%V',strtotime("2017-01-01"))
Result: 2016-52
https://www.php.net/manual/de/function.strftime.php
ISO-8601:1988 week number of the given year, starting with the first week of the year with at least 4 weekdays, with Monday being the start of the week. (01 through 53)
The equivalent in mysql is DATE_FORMAT(date, '%x-%v')
https://www.w3schools.com/sql/func_mysql_date_format.asp
Week where Monday is the first day of the week (01 to 53).
Could not find a corresponding solution with DateTime.
At least not without solutions like "+1day, last monday".
Edit: since strftime is now deprecated, maybe you can also use date.
Didn't verify it though.
date('o-W',strtotime("2017-01-01"));
I have tried to solve this question for years now, I thought I found a shorter solution but had to come back again to the long story. This function gives back the right ISO week notation:
/**
* calcweek("2018-12-31") => 1901
* This function calculates the production weeknumber according to the start on
* monday and with at least 4 days in the new year. Given that the $date has
* the following format Y-m-d then the outcome is and integer.
*
* #author M.S.B. Bachus
*
* #param date-notation PHP "Y-m-d" showing the data as yyyy-mm-dd
* #return integer
**/
function calcweek($date) {
// 1. Convert input to $year, $month, $day
$dateset = strtotime($date);
$year = date("Y", $dateset);
$month = date("m", $dateset);
$day = date("d", $dateset);
$referenceday = getdate(mktime(0,0,0, $month, $day, $year));
$jan1day = getdate(mktime(0,0,0,1,1,$referenceday[year]));
// 2. check if $year is a leapyear
if ( ($year%4==0 && $year%100!=0) || $year%400==0) {
$leapyear = true;
} else {
$leapyear = false;
}
// 3. check if $year-1 is a leapyear
if ( (($year-1)%4==0 && ($year-1)%100!=0) || ($year-1)%400==0 ) {
$leapyearprev = true;
} else {
$leapyearprev = false;
}
// 4. find the dayofyearnumber for y m d
$mnth = array(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334);
$dayofyearnumber = $day + $mnth[$month-1];
if ( $leapyear && $month > 2 ) { $dayofyearnumber++; }
// 5. find the jan1weekday for y (monday=1, sunday=7)
$yy = ($year-1)%100;
$c = ($year-1) - $yy;
$g = $yy + intval($yy/4);
$jan1weekday = 1+((((intval($c/100)%4)*5)+$g)%7);
// 6. find the weekday for y m d
$h = $dayofyearnumber + ($jan1weekday-1);
$weekday = 1+(($h-1)%7);
// 7. find if y m d falls in yearnumber y-1, weeknumber 52 or 53
$foundweeknum = false;
if ( $dayofyearnumber <= (8-$jan1weekday) && $jan1weekday > 4 ) {
$yearnumber = $year - 1;
if ( $jan1weekday = 5 || ( $jan1weekday = 6 && $leapyearprev )) {
$weeknumber = 53;
} else {
$weeknumber = 52;
}
$foundweeknum = true;
} else {
$yearnumber = $year;
}
// 8. find if y m d falls in yearnumber y+1, weeknumber 1
if ( $yearnumber == $year && !$foundweeknum) {
if ( $leapyear ) {
$i = 366;
} else {
$i = 365;
}
if ( ($i - $dayofyearnumber) < (4 - $weekday) ) {
$yearnumber = $year + 1;
$weeknumber = 1;
$foundweeknum = true;
}
}
// 9. find if y m d falls in yearnumber y, weeknumber 1 through 53
if ( $yearnumber == $year && !$foundweeknum ) {
$j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
$weeknumber = intval( $j/7 );
if ( $jan1weekday > 4 ) { $weeknumber--; }
}
// 10. output iso week number (YYWW)
return ($yearnumber-2000)*100+$weeknumber;
}
I found out that my short solution missed the 2018-12-31 as it gave back 1801 instead of 1901. So I had to put in this long version which is correct.
How about using the IntlGregorianCalendar class?
Requirements: Before you start to use IntlGregorianCalendar make sure that libicu or pecl/intl is installed on the Server.
So run on the CLI:
php -m
If you see intl in the [PHP Modules] list, then you can use IntlGregorianCalendar.
DateTime vs IntlGregorianCalendar:
IntlGregorianCalendar is not better then DateTime. But the good thing about IntlGregorianCalendar is that it will give you the week number as an int.
Example:
$dateTime = new DateTime('21-09-2020 09:00:00');
echo $dateTime->format("W"); // string '39'
$intlCalendar = IntlCalendar::fromDateTime ('21-09-2020 09:00:00');
echo $intlCalendar->get(IntlCalendar::FIELD_WEEK_OF_YEAR); // integer 39
<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>
You had the params to mktime wrong - needs to be Month/Day/Year, not Day/Month/Year
To get the week number for a date in North America I do like this:
function week_number($n)
{
$w = date('w', $n);
return 1 + date('z', $n + (6 - $w) * 24 * 3600) / 7;
}
$n = strtotime('2022-12-27');
printf("%s: %d\n", date('D Y-m-d', $n), week_number($n));
and get:
Tue 2022-12-27: 53
for get week number in jalai calendar you can use this:
$weeknumber = date("W"); //number week in year
$dayweek = date("w"); //number day in week
if ($dayweek == "6")
{
$weeknumberint = (int)$weeknumber;
$date2int++;
$weeknumber = (string)$date2int;
}
echo $date2;
result:
15
week number change in saturday
The most of the above given examples create a problem when a year has 53 weeks (like 2020). So every fourth year you will experience a week difference. This code does not:
$thisYear = "2020";
$thisDate = "2020-04-24"; //or any other custom date
$weeknr = date("W", strtotime($thisDate)); //when you want the weeknumber of a specific week, or just enter the weeknumber yourself
$tempDatum = new DateTime();
$tempDatum->setISODate($thisYear, $weeknr);
$tempDatum_start = $tempDatum->format('Y-m-d');
$tempDatum->setISODate($thisYear, $weeknr, 7);
$tempDatum_end = $tempDatum->format('Y-m-d');
echo $tempDatum_start //will output the date of monday
echo $tempDatum_end // will output the date of sunday
Very simple
Just one line:
<?php $date=date("W"); echo "Week " . $date; ?>"
You can also, for example like I needed for a graph, subtract to get the previous week like:
<?php $date=date("W"); echo "Week " . ($date - 1); ?>
Your code will work but you need to flip the 4th and the 5th argument.
I would do it this way
$date_string = "2012-10-18";
$date_int = strtotime($date_string);
$date_date = date($date_int);
$week_number = date('W', $date_date);
echo "Weeknumber: {$week_number}.";
Also, your variable names will be confusing to you after a week of not looking at that code, you should consider reading http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/
The rule is that the first week of a year is the week that contains the first Thursday of the year.
I personally use Zend_Date for this kind of calculation and to get the week for today is this simple. They have a lot of other useful functions if you work with dates.
$now = Zend_Date::now();
$week = $now->get(Zend_Date::WEEK);
// 10
To get Correct Week Count for Date 2018-12-31 Please use below Code
$day_count = date('N',strtotime('2018-12-31'));
$week_count = date('W',strtotime('2018-12-31'));
if($week_count=='01' && date('m',strtotime('2018-12-31'))==12){
$yr_count = date('y',strtotime('2018-12-31')) + 1;
}else{
$yr_count = date('y',strtotime('2018-12-31'));
}
function last_monday($date)
{
if (!is_numeric($date))
$date = strtotime($date);
if (date('w', $date) == 1)
return $date;
else
return date('Y-m-d',strtotime('last monday',$date));
}
$date = '2021-01-04'; //Enter custom date
$year = date('Y',strtotime($date));
$date1 = new DateTime($date);
$ldate = last_monday($year."-01-01");
$date2 = new DateTime($ldate);
$diff = $date2->diff($date1)->format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.
try this solution
date( 'W', strtotime( "2017-01-01 + 1 day" ) );