我有一个返回1个月前网址的函数。
我想显示当前选定的月份，但我不能使用简单的当月，因为当用户点击链接到1个月后，所选月份将会改变并且不会是最新的。
因此，函数将于2012年8月返回
如何制作一个增加1个月的小PHP脚本？
到目前为止我已经：
<?php echo strip_tags(tribe_get_previous_month_text()); ?>
答案 0 :(得分：9)
简单方法：
$next_month = strtotime('august 2012 next month');
更好的方法：
$d = new Date('August 2012');
$next_month = $d->add(new DateInterval('P1M'));
相关文档：strtotime date dateinterval
答案 1 :(得分：2)
您可以使用DateTime类和DateTime :: add（）方法：
答案 2 :(得分：2)
有3个选项/答案
$givendate is the given date (ex. 2016-01-20)
option 1:
$date1 = date('Y-m-d', strtotime($givendate. ' + 1 month'));
option 2:
$date2 = date('Y-m-d', strtotime($givendate. ' + 30 days'));
option 3:
$number = cal_days_in_month(CAL_GREGORIAN, date('m', strtotime($givendate)), date('Y', strtotime($givendate)));
$date3 = date('Y-m-d', strtotime($date2. ' + '.$number.' days'));
答案 3 :(得分：1)
您可以简单地使用strtotime功能对您必须在2012年4月到达的输入进行操作，然后应用date和strtotime，增量周期为＆lt; + 1; + 1个月＆ ＃39 ;.
$x = strtotime($t);
$n = date("M Y",strtotime("+1 month",$x));
echo $n;
以下是PHP手册的相关章节：
http://www.php.net/manual/en/function.date.php
https://secure.php.net/manual/en/function.strtotime.php
此解决方案解决了将任何时间量增加到时间值的其他问题。
答案 4 :(得分：0)
您好，除了他们的回答。我想，如果您只想根据当前日期获得下个月，这就是我的解决方法。
$today = date("Y-m-01");
$sNextMonth = (int)date("m",strtotime($today." +1 months") );
注意，我不断将日期定义为01，以便我们可以安全地获取下个月。如果是日期（“ Y-m-d”）;而今天是31天，它将失败。
希望这会有所帮助。
答案 5 :(得分：0)
日期差异
$date1 = '2017-01-20';
$date2 = '2019-01-20';
$ts1 = strtotime($date1);
$ts2 = strtotime($date2);
$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);
$month1 = date('m', $ts1);
$month2 = date('m', $ts2);
echo $joining_months = (($year2 - $year1) * 12) + ($month2 - $month1);
答案 6 :(得分：-1)
因为我们知道 strtotime(+1 个月) 总是增加 30 天，所以如果日期以第 31、30 或 29 天结束，并且如果您仍然想留在下个月的最后一天，则可能会遇到一些麻烦。< /p>
所以我写了这个复杂的脚本来解决这个问题并进行调整，以便您可以增加所有类型的格式，如年、月、日、小时、分钟和秒。
function seetime($datetime, $p = '+', $i, $m = 'M', $f = 'Y-m-d H:i:s')
{
/*
$datetime needs to be in format of YYYY-MM-DD HH:II:SS but hours, minutes and seconds are not required
$p can only be "+" to increse or "-" to decrese
$i is the amount you want to change
$m is the type you want to change
Allowed types:
Y = Year
M = Months
D = Days
W = Weeks
H = Hours
I = Minutes
S = Seconds
$f is the datetime format you want the result to be returned in
*/
$validator_y = substr($datetime,0,4);
$validator_m = substr($datetime,5,2);
$validator_d = substr($datetime,8,2);
if(checkdate($validator_m, $validator_d, $validator_y))
{
$datetime = date('Y-m-d H:i:s', strtotime($datetime));
#$p = either "+" to add or "-" to subtract
if($p == '+' || $p == '-')
{
if(is_int($i))
{
if($m == 'Y')
{
$year = date('Y', strtotime($datetime));
$rest = date('m-d H:i:s', strtotime($datetime));
if($p == '+')
{
$ret = $year + $i;
}
else
{
$ret = $year - $i;
}
$str = $ret.'-'.$rest;
return(date($f, strtotime($str)));
}
elseif($m == 'M')
{
$year = date('Y', strtotime($datetime));
$month = date('n', strtotime($datetime));
$rest = date('d H:i:s', strtotime($datetime));
$his = date('H:i:s', strtotime($datetime));
if($p == '+')
{
$ret = $month + $i;
$ret = sprintf("%02d",$ret);
}
else
{
$ret = $month - $i;
$ret = sprintf("%02d",$ret);
}
if($ret < 1)
{
$ret = $ret - $ret - $ret;
$years_back = floor(($ret + 12) / 12);
$monts_back = $ret % 12;
$year = $year - $years_back;
$month = 12 - $monts_back;
$month = sprintf("%02d",$month);
$new_date = $year.'-'.$month.'-'.$rest;
$ym = $year.'-'.$month;
$validator_y = substr($new_date,0,4);
$validator_m = substr($new_date,5,2);
$validator_d = substr($new_date,8,2);
if(checkdate($validator_m, $validator_d, $validator_y))
{
return (date($f, strtotime($new_date)));
}
else
{
$days = date('t',strtotime($ym));
$new_date = $ym.'-'.$days.' '.$his;
return (date($f, strtotime($new_date)));
}
}
if($ret > 12)
{
$years_forw = floor($ret / 12);
$monts_forw = $ret % 12;
$year = $year + $years_forw;
$month = sprintf("%02d",$monts_forw);
$new_date = $year.'-'.$month.'-'.$rest;
$ym = $year.'-'.$month;
$validator_y = substr($new_date,0,4);
$validator_m = substr($new_date,5,2);
$validator_d = substr($new_date,8,2);
if(checkdate($validator_m, $validator_d, $validator_y))
{
return (date($f, strtotime($new_date)));
}
else
{
$days = date('t',strtotime($ym));
$new_date = $ym.'-'.$days.' '.$his;
return (date($f, strtotime($new_date)));
}
}
else
{
$ym = $year.'-'.$month;
$new_date = $year.'-'.$ret.'-'.$rest;
$validator_y = substr($new_date,0,4);
$validator_m = substr($new_date,5,2);
$validator_d = substr($new_date,8,2);
if(checkdate($validator_m, $validator_d, $validator_y))
{
return (date($f, strtotime($new_date)));
}
else
{
$ym = $validator_y . '-'.$validator_m;
$days = date('t',strtotime($ym));
$new_date = $ym.'-'.$days.' '.$his;
return (date($f, strtotime($new_date)));
}
}
}
elseif($m == 'D')
{
return (date($f, strtotime($datetime.' '.$p.$i.' days')));
}
elseif($m == 'W')
{
return (date($f, strtotime($datetime.' '.$p.$i.' weeks')));
}
elseif($m == 'H')
{
return (date($f, strtotime($datetime.' '.$p.$i.' hours')));
}
elseif($m == 'I')
{
return (date($f, strtotime($datetime.' '.$p.$i.' minutes')));
}
elseif($m == 'S')
{
return (date($f, strtotime($datetime.' '.$p.$i.' seconds')));
}
else
{
return 'Fourth parameter can only be any of following: Valid Time Parameters Are: Y M D Q H I S';
}
}
else
{
return 'Third parameter can only be a number (whole number)';
}
}
else
{
return 'Second parameter can only be + to add or - to subtract';
}
}
else
{
return 'Date is not a valid date';
}
}