日付の配列から特定の月だけ指定して日付だけ抽出したい

Question

内閣府ホームページの"国民の祝日から祝日データーを抽出しました。
$horidays
には
Array ( [0] => 2019/1/1 [1] => 2019/1/14 [2] => 2019/2/11 [3] => 2019/3/21 [4] => 2019/4/29 [5] => 2019/4/30 [6] => 2019/5/1 [7] => 2019/5/2 [8] => 2019/5/3 [9] => 2019/5/4 [10] => 2019/5/5 [11] => 2019/5/6 [12] => 2019/7/15 [13] => 2019/8/11 [14] => 2019/8/12 [15] => 2019/9/16 [16] => 2019/9/23 [17] => 2019/10/14 [18] => 2019/10/22 [19] => 2019/11/3 [20] => 2019/11/4 [21] => 2019/11/23 [22] => 2020/1/1 [23] => 2020/1/13 [24] => 2020/2/11 [25] => 2020/2/23 [26] => 2020/2/24 [27] => 2020/3/20 [28] => 2020/4/29 [29] => 2020/5/3 [30] => 2020/5/4 [31] => 2020/5/5 [32] => 2020/5/6 [33] => 2020/7/23 [34] => 2020/7/24 [35] => 2020/8/10 [36] => 2020/9/21 [37] => 2020/9/22 [38] => 2020/11/3 [39] => 2020/11/23 )

上記のように日付が格納されています。

ここから５月だけを抜き取って
3,4,5,6
と日付だけを表示させたいのですがなにか方法はあるでしょうか？？

よろしくお願いします。

Answer 1

$horidays=["2019/1/1","2019/1/14","2019/2/11","2019/3/21","2019/4/29","2019/4/30","2019/5/1","2019/5/2","2019/5/3","2019/5/4","2019/5/5","2019/5/6","2019/7/15","2019/8/11","2019/8/12","2019/9/16","2019/9/23","2019/10/14","2019/10/22","2019/11/3","2019/11/4","2019/11/23","2020/1/1","2020/1/13","2020/2/11","2020/2/23","2020/2/24","2020/3/20","2020/4/29","2020/5/3","2020/5/4","2020/5/5","2020/5/6","2020/7/23","2020/7/24","2020/8/10","2020/9/21","2020/9/22","2020/11/3","2020/11/23"];

$year=2020;
$month=5;

print implode(",",array_map(function($x){
  return date("j",strtotime($x));
},array_filter($horidays,function($x) use($year,$month){
  return date("Y",strtotime($x))==$year and date("m",strtotime($x))==$month;
})));

Comments

[kazu130]

返信ありがとうございます。こちらでも結果がでました。

Answer 2

愚直ですが、ループで回して月判定すればよろしいかと。

foreach($horidays as $v){
    if(date("m", strtotime($v)) == 5){
        printf("%d,", date("d", strtotime($v)));
    }
}

余談ですが、5/1と5/2は除外しなくていいんですよね？

---

単純に/で分割して判定でいいような気がするのでこちらを本回答とします。

foreach($horidays as $v){
    list($y,$m,$d) = explode("/", $v);
    if($m == 5){
        printf("%d,", $d);
    }
}

年の判定も必要ならこう。

foreach($horidays as $v){
    list($y,$m,$d) = explode("/", $v);
    if($y == 2020 && $m == 5){
        printf("%d,", $d);
    }
}

Comments

[kazu130]

遅くなって申し訳ございません。
はい。1,2日除外してましたが必要です。
狙い通り行けました。ありがとうございます。