如何在php中获取每个月的最后一个日期，年份将根据cuurent日期而变化

我想得到每年每个月的最后一个日期。年份将发生变化，从currdate开始计算

我试过低于一

$day = date( 't-m-Y' );

但是我每个月都会写不同的if条件。所以在这段代码中，我想给出常量月。这一年将会改变。有可能这样做吗

有人能帮我吗。

检查演示

<?php
$year = date("Y");
echo $year . "\n";
for ($i = 1; $i <= 12; $i++)
{
    $month = strtotime("$year-$i");
    echo date("t", $month) . "\n";
}

---

你可以检查这个代码

$a_date = "2017-04-17";
echo date("Y-m-t", strtotime($a_date));

或使用

$current_date = "2017-04-17";
$date = new DateTime($current_date);
$date->modify('last day of this month');
echo $date->format('Y-m-d');


echo $first_date = date('Y-m-d',strtotime('first day of this month'));
echo $last_date = date('Y-m-d',strtotime('last day of this month'));

这是你的密码

echo $last_date = date('t-03-Y');

这就是输出

30-03-2017

---

t

返回给定日期当月的天数

所以你可以像这样通过动态年份和固定月份来知道月份的最后一天

<?php

 $year ="2017";
 echo date("Y-m-t", strtotime(date("$year-04-d")));

?>

更新1:

2017-04-30

 $date = new DateTime('last day of 2017-04');

 echo $date->format('Y-m-d');

---

月份的最后一天年复一年保持不变，即12月将有31天，无论是指今年还是下一年。唯一的变化是在一周的哪一天，一个月的最后一天。因此，下面说明了如何获取当前月份、今年剩余月份以及明年的所有这些数据

<?php

// preliminaries
try {
    $date =  new DateTime("@".time());
} catch (Exception $e) {
    echo $e->getMessage();
    exit(1);
}

$strDate = $date->format("Y-n-t");
list($Y,$m,$t)  =  explode("-",$strDate);
$offset = 12 - $m;
$max = $offset + 1; // include current month


//this year starting with current month:
for ( $i=0; $i < $max; $i++ ){
    $mo_day_one_yr = "$Y-" . ( $m + $i ) . "-1";
    $date->modify( $mo_day_one_yr );
    echo $date->format("D, t-M-Y"),"\n";
    if( $i == ( $max -1 )) echo "\n";
}


// next year:
for ( $i=1, $Y+=1, $max=12; $i <= $max; $i++ ){
    $mo_day_one_yr = "$Y-$i-1";
    $date->modify( $mo_day_one_yr );
    echo $date->format("D, t-M-Y"),"\n";
}

---

t返回给定日期当月的天数，请尝试此$date=“2009-11-23”；回音日期（“t-m-y”，标准时间（$date））；我们可以这样用吗。日期（'t-03-Y'）？？因为在每个if条件下，我都知道月份，我只想得到今年和未来该月的最后日期。当然，您可以根据自己的目的编辑此代码。
// get Current Date 
$date = date('j-m-Y'); 
echo date("j-m-y", strtotime($date));//Current Month date with Current year
echo "Current Year =".date("Y", strtotime($date))." , Current Month Days =".date("t", strtotime($date))." , Current Month Last Day Date  =".date("t-m-Y", strtotime($date));

<?php

// preliminaries
try {
    $date =  new DateTime("@".time());
} catch (Exception $e) {
    echo $e->getMessage();
    exit(1);
}

$strDate = $date->format("Y-n-t");
list($Y,$m,$t)  =  explode("-",$strDate);
$offset = 12 - $m;
$max = $offset + 1; // include current month


//this year starting with current month:
for ( $i=0; $i < $max; $i++ ){
    $mo_day_one_yr = "$Y-" . ( $m + $i ) . "-1";
    $date->modify( $mo_day_one_yr );
    echo $date->format("D, t-M-Y"),"\n";
    if( $i == ( $max -1 )) echo "\n";
}


// next year:
for ( $i=1, $Y+=1, $max=12; $i <= $max; $i++ ){
    $mo_day_one_yr = "$Y-$i-1";
    $date->modify( $mo_day_one_yr );
    echo $date->format("D, t-M-Y"),"\n";
}