how to create array of a week days name in php

Question

I know this is stupid, but how would I do this...? i want to create an array of 7 days through php means all 7 days name... i don't want to write ..sunday monday tuesday...etc... and days will be started from sunday .. means if today is 29 march( monday) then it automatically grabs the current date and create an array of days name starting from sunday. array always be in this way $weakarray=("sunday","monday",......,"saturday");

Answer 1

$now = time();
$days = array();
for ($i = 0; $i < 7; $i++) {
    $days[] = strftime('%A', $now);
    $now += 60*60*24;
}

Answer 2

If they always need to start with Sunday why do you want to create the array dynamically? What is wrong with doing this?

$days = array(
    'Sunday',
    'Monday',
    'Tuesday',
    'Wednesday',
    'Thursday',
    'Friday',
    'Saturday',
);

Any other solution is going to make your code harder to understand and in this case doing it dynamically seems to be overkill.

Answer 3

This might work..

$timestamp = strtotime('next Sunday');
$days = array();
for ($i = 0; $i < 7; $i++) {
 $days[] = strftime('%A', $timestamp);
 $timestamp = strtotime('+1 day', $timestamp);
}

Answer 4

Try:

for($i=1;$i<8;$i++)
$weekdays [] = date("l",mktime(0,0,0,3,28,2009)+$i * (3600*24));
var_dump($weekdays);

Output:

C:\>php b.php
array(7) {
  [0]=>
  string(6) "Sunday"
  [1]=>
  string(6) "Monday"
  [2]=>
  string(7) "Tuesday"
  [3]=>
  string(9) "Wednesday"
  [4]=>
  string(8) "Thursday"
  [5]=>
  string(6) "Friday"
  [6]=>
  string(8) "Saturday"
}

Answer 5

$days = array();
for ($x = 0; $x < 7; $x++) {
    $days[] = date('l', strtotime("+$x days", strtotime('2010-03-28')));
}

Seriously though, unless your question is being misunderstood, I completely second Yacoby's answer.

Answer 6

use the DateTime object

$date = new DateTime();
$weekdays = array();
for($i=0; $i<7; $i++){
    $weekdays[] = $date->format('l');
    $date->modify('+1 day');
}

If you want to start with Sunday, create DateTime with a sunday day (for example, yesterday):

$date = new DateTime('2010-03-28');