weeks in a year


Posted on 16th Feb 2014 07:03 pm by admin

hi,

i found this snippet on php.net
QuoteFor the week number for weeks starting on Sunday:
Did you know?Explore Trending and Topic pages for more stories like this.

<?php
function week_of_year($month, $day, $year) {

$day_of_year = date('z', mktime(0, 0, 0, $month, $day, $year));

/* Days in the week before Jan 1. If you want weeks to start on Monday make this (x + 6) % 7 */
$days_before_year = date('w', mktime(0, 0, 0, 1, 1, $year));

$days_left_in_week = 7 - date('w', mktime(0, 0, 0, $month, $day, $year));

/* find the number of weeks by adding the days in the week before the start of the year, days up to $day, and the days left in this week, then divide by 7 */
return ($days_before_year + $day_of_year + $days_left_in_week) / 7;

}
?>


so if i go : week_of_year(1, 3, 2010) (sunday)

the week shows as 1, because the script is set to begin new week on sunday,

however i am confused by his instruction " (x + 6) % 7 " to have the week start on a monday

what part of the script do i apply this equation ???

thanks
No comments posted yet

Your Answer:

Login to answer
142 Like 13 Dislike
Previous forums Next forums
Other forums

Delete HTML file after loading
I have limited experience with php and its been a year or two since I've last used it. I have a sma

Module pool selction screen parameters combination logic
Hi floks,
Am new to the module pool development ,Recently i have created one program based on

Please help - should be a simple fix.. driving me nuts
Everything seemed to be working fine. I have a table, it alphabetically lists a bunch of cities and

Time-based image rotation script
I'm trying to write a PHP script that rotates an image based on what time of day it is. I want the

Calander layout
Hi i know this sounds like a simple question but i cant find the answer to it anywhere i have added

Creating Images from images in PHP
Okay well I am trying to make a 'dynamic' calender image with PHP.

I have images like this:

Form validation with functions
Hi there

I am trying to make a very simple form validation function. I currently have the fol

Ajax Error since Upgrading to 3.5
Ever since upgrading my site to .NET 3.5 (I needed LINQ), I've been getting this annoying error on o

PHP using IF to display error
i have a MySQL query and i want to display 1 thing only if the number of affected rows is >=1

images aren't rendering
I'm trying to call a JPG file from within PHP (in an effort to hide the actual JPG folder). The imag

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash