php - 2 つの日付の範囲から平日の日付を検索する

Question

$start_date = "2013-05-01";
$last_date = "2013-08-30";

これら2つの日付の間の火曜日と木曜日の日付を取得するにはどうすればよいですか?

Answer 1

一部の PHP-Fu

$start_date = '2013-05-01';
$last_date = '2013-08-30';

$dates = range(strtotime($start_date), strtotime($last_date),86400);
$days = array('tuesday' => array(), 'thursday' => array());

array_map(function($v)use(&$days){
        if(date('D', $v) == 'Tue'){
            $days['tuesday'][] = date('Y-m-d', $v);
        }elseif(date('D', $v) == 'Thu'){
            $days['thursday'][] = date('Y-m-d', $v);
        }
    }, $dates); // Requires PHP 5.3+

print_r($days);

出力

Array
(
    [tuesday] => Array
        (
            [0] => 2013-05-07
            [1] => 2013-05-14
            [2] => 2013-05-21
            [3] => 2013-05-28
            [4] => 2013-06-04
            [5] => 2013-06-11
            [6] => 2013-06-18
            [7] => 2013-06-25
            [8] => 2013-07-02
            [9] => 2013-07-09
            [10] => 2013-07-16
            [11] => 2013-07-23
            [12] => 2013-07-30
            [13] => 2013-08-06
            [14] => 2013-08-13
            [15] => 2013-08-20
            [16] => 2013-08-27
        )

    [thursday] => Array
        (
            [0] => 2013-05-02
            [1] => 2013-05-09
            [2] => 2013-05-16
            [3] => 2013-05-23
            [4] => 2013-05-30
            [5] => 2013-06-06
            [6] => 2013-06-13
            [7] => 2013-06-20
            [8] => 2013-06-27
            [9] => 2013-07-04
            [10] => 2013-07-11
            [11] => 2013-07-18
            [12] => 2013-07-25
            [13] => 2013-08-01
            [14] => 2013-08-08
            [15] => 2013-08-15
            [16] => 2013-08-22
            [17] => 2013-08-29
        )

)

オンラインデモ

Answer 2

いくつかの php 日付関数の助けを借りて、これは簡単に解決できます..

<?php

// Create the from and to date
$start_date = strtotime("2013-05-01");
$last_date  = strtotime("2013-08-30");

// Get the time interval to get the tue and Thurs days
$no_of_days = ($last_date - $start_date) / 86400; //the diff will be in timestamp hence dividing by timestamp for one day = 86400
$get_tue_thu_days = array();

// Loop upto the $no_of_days
for($i = 0; $i < $no_of_days; $i++) {
    $temp = date("D", $start_date);
    if($temp == "Tue" || $temp == "Thu") {
      $get_tue_thu_days[] = date("D/M/Y", $start_date); //formating date in Thu/May/2013 formate.
    }
    $start_date += 86400;
}

print_r($get_tue_thu_days);

Answer 3

DateTimeの場合:

$start_date = "2013-05-01";
$last_date = "2013-08-30";

$start = new DateTime($start_date);
$clone = clone $start;

$start->modify('next thursday');
$thursday=$start->format('Y-m-d');

$clone->modify('next tuesday');
$tuesday=$clone->format('Y-m-d');

echo $thursday; //2013-05-02
echo $tuesday; //2013-05-07

tuesdayin intervalが beforeの場合、 thursdaynextがあるため、オブジェクトが必要ですtuesday。ただし、わずかなコードを変更して 1 つのオブジェクトを使用することはできます。

Answer 4

基準日が火曜日/木曜日であることがわかっている場合、基準日から 7 日の倍数である日を見つけることができます。これらの日は常に同じ曜日になります。