Joe*_*oel 95
尝试:
date("Y-m-d", strtotime("+1 week"));
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这将输出:
2015-12-31
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如果今天是2015-12-24
Mar*_*c B 19
正如查尔斯的预测错误,这是一个PHP 5.3+示例:
$now = new DateTime;
$interval = new DateInterval('P1W')
$next_week = $now->add($interval);
echo $next_week->format('Y-m-d');
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或者稍微紧凑的形式:
$now = new DateTime();
echo $now->add(new DateInterval('P1W'))->format('Y-m-d');
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<?php
$nextWeek = time() + (7 * 24 * 60 * 60);
echo 'Next Week: '. date('Y-m-d', $nextWeek) ."\n";
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查尔斯预测的一个人,直接来自马口,例如#1
将天,周,月添加到任何日期
$date = date("Y-m-d");// current date
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 day");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +2 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 month");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days");
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