使用未定义的常量 ? - 假设'？' 在 php 中
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Use of undefined constant ? - assumed '?' in php
提问by PHP Learner
<?php
$f=$_POST['rdate'];
echo $f."</br>";
$from=$f.'-01 00:00:00';
$dayyy=explode("-",$f);
//print_r($dayyy);
$year=$dayyy[0];
echo $year."</br>";
$mm=$dayyy[1];
echo $mm."</br>";
$mons = array("01" => "Janauary", "02" => "February", "03"=>"March", "04"=>"April", "05"=>"May", "06"=>"June", "07"=>"July", "08"=>"August", "09"=>"September", "10"=>"October", "11"=>"November", "12"=>"December");
print_r($mons);
/*foreach($mons as $mm)
{
echo $mm;
}*/
$month_name = $mons[$mm];
echo "</br>".$month_name;
$days=cal_days_in_month(CAL_GREGORIAN, $mm, $year);
$dayss= $days-1;
echo $dayss."</br>";
$dayys=' + '.$dayss.' days' ;
$var=$from.$dayys;
//echo $var."</br>";
echo "Last".date('Y-m-d 23:59:59',strtotime($var));?
?>
i'm getting Use of undefined constant ? - assumed error in the above code. I guess i got this issue due to non breaking spce but don't know where i left that? Kindly help!!
我正在使用未定义的常量？- 上面代码中的假设错误。我想我是因为没有破坏 spce 而遇到这个问题的，但不知道我把它放在哪里了？请帮忙！！
回答by dano
You have some kind of weird character on your
你身上有某种奇怪的性格
echo "Last".date('Y-m-d 23:59:59',strtotime($var));
line notepad++ couldn't see it but when I tried to arrow throw it, it went through twice.
行记事本++看不到它，但是当我尝试用箭头扔它时，它经历了两次。
Anyways, remove the line and copy and paste this
无论如何，删除该行并复制并粘贴
echo "Last".date('Y-m-d 23:59:59',strtotime($var));
回答by Azeem Hassni
check the constant you are using is defined or not .. if not define it
检查您正在使用的常量是否已定义.. 如果没有定义它
the code will be
代码将是
<?php
if (!defined('CAL_GREGORIAN'))
define('CAL_GREGORIAN', 0);
$f=$_POST['rdate'];
echo $f."</br>";
$from=$f.'-01 00:00:00';
$dayyy=explode("-",$f);
//print_r($dayyy);
$year=$dayyy[0];
echo $year."</br>";
$mm=$dayyy[1];
echo $mm."</br>";
$mons = array("01" => "Janauary", "02" => "February", "03"=>"March", "04"=>"April", "05"=>"May", "06"=>"June", "07"=>"July", "08"=>"August", "09"=>"September", "10"=>"October", "11"=>"November", "12"=>"December");
print_r($mons);
/*foreach($mons as $mm)
{
echo $mm;
}*/
$month_name = $mons[$mm];
echo "</br>".$month_name;
$days=cal_days_in_month(CAL_GREGORIAN, $mm, $year);
$dayss= $days-1;
echo $dayss."</br>";
$dayys=' + '.$dayss.' days' ;
$var=$from.$dayys;
//echo $var."</br>";
echo "Last".date('Y-m-d 23:59:59',strtotime($var));
