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TA貢獻1828條經驗 獲得超4個贊
一個簡單的 IF 應該允許您不打印日期之一
$aDay = 86400*3;
$date = time() + $aDay;
$b = 5; // Anzahl Schleifendurchl?ufe
$cb = "4";
echo 'Loop without any checks'.PHP_EOL;
for($i = 0; $i < $b; $i++) {
echo date('d.m.Y',$date) . PHP_EOL;
$date += $aDay;
}
//reset date and pick one of the dates that will occur and stop it
$date = time() + $aDay;
echo PHP_EOL.'Loop with checks'.PHP_EOL;
for($i = 0; $i < $b; $i++) {
$cb++;
if (date('d.m', $date) == '07.01'){
echo 'Dont print it ' . date('d.m.Y', $date) . PHP_EOL;
}else{
echo 'Print it ' .date('d.m.Y', $date) . PHP_EOL;
}
$date += $aDay;
}
結果
Loop without any checks
01.01.2020
04.01.2020
07.01.2020
10.01.2020
13.01.2020
Loop with checks
Print it 01.01.2020
Print it 04.01.2020
Dont print it 07.01.2020
Print it 10.01.2020
Print it 13.01.2020
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