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Numerical date to string conversion

开发者 https://www.devze.com 2023-02-20 13:56 出处:网络
I am using the following code to convert a date in the format mm/dd/yy to show November, 19, 2008 echo date(\'F d, Y\', strtotime(\'{$month}/{$day}/{$year}\'));

I am using the following code to convert a date in the format mm/dd/yy to show November, 19, 2008

echo date('F d, Y', strtotime('{$month}/{$day}/{$year}'));

In my database I have the values for $month, $day and $year 开发者_开发百科stored as an INT.

When I add the variables into the date function, the output is wrong. I am getting an unexpected date. The date I have in the system is 2/1/1977 but the output gives December 31, 1969.

any ideas?


I think the problem isyour single quotes around the string in strtotime

Single quotes wont parse and put in the variables value, it will be exactly that literal. you want to use doubles

$month = 1; $day = 2; $year = 2000;
echo date('F d, Y', strtotime("$month/$day/$year")); 
echo '    {$month}/{$day}/{$year}';  //gives exactly whats between the single quotes

Output :

January 02, 2000    {$month}/{$day}/{$year}

EDIT

for($=0;$i<count($month);$i++)
   echo date('F d, Y', strtotime($month[$i].'/'.$day[$i].'/'.$year[$i])) . '<br />'; 
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