Вопрос

Goal: If current day of week is any day other than Monday, display the date of the Monday of the current week. If the current day of the week is Monday, simply display today's date.

** This is what I wrote and I think it works but is probably not the cleanest way to determine the date. Having said that, does anyone see any reason why the code would be wrong or not work? **

<?php
date_default_timezone_set("America/New_York");
$day = date("w");
if( $day == 1 ) {$day -= 0;}
if( $day == 2 ) {$day -= 1;}
if( $day == 3 ) {$day -= 2;}
if( $day == 4 ) {$day -= 3;}
if( $day == 5 ) {$day -= 4;}
if( $day == 6 ) {$day -= 5;}
if( $day == 0 ) {$day -= 6;}
$calc = mktime(0,0,0,date("m"),date("d")-$day,date("Y"));
echo date("m/d/Y",$calc)."<br>";
echo "start date of work week is: ".date("m/d/Y", $calc);
?>
Это было полезно?

Решение

wont this work?

echo date("d m Y",strtotime('monday this week'));

Другие советы

You Can try This :

date_default_timezone_set("Asia/Kolkata"); $day = date("w")."<br>"; echo $day; //if( $day == 1 ) {$day = 0;} if( $day == 2 ) {$day -= 1;} if( $day == 3 ) {$day -= 1;} if( $day == 4 ) {$day -= 1;} if( $day == 5 ) {$day -= 1;} if( $day == 6 ) {$day -= 1;} if( $day == 0 ) {$day -= 6;} $calc = mktime(0,0,0,date("m"),date("d")-$day,date("Y")); if($day == 1 ){ echo date("m/d/Y")."<br>"; echo "start date of work week is: ".date("m/d/Y"); } else{ echo date("m/d/Y",$calc)."<br>"; echo "start date of work week is: ".date("m/d/Y", $calc); }
$time = strtotime('monday this week');

The following works if your week starts on Monday.

$monday = strtotime('this sunday', strtotime($startingDate)) - (86400 * 6);

While if you do it the way @mithunsatheesh suggested works perfectly if your week starts on Sunday.

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