在PHP中将前导零添加到字符串日期

Question

I have a string "date" which can be DD.MM.YYYY or D.M.YYYY (with or without leading zeros), it depends what a user types. Then I use it in a condition to send another email when the day is today.

if($_POST["date"]== date("d.m.Y")){ 
  $headers.="Bcc: [e-mail]
";
}

The problem is that the mail is send when the date format DD.MM.YYYY (with leading zeros) only.

My proposed solution

As I'm not very good in PHP, I only know the solution theoretically but not how to write the code - I would spend a week trying to figure it out on my own.

What's in my mind is dividing the date into three parts (day, month, year), then checking the first two parts if there's just one digit and adding leading zeros if it's the case. I don't know how to implement that to the condition above, though. I have read a few topics about how to do this, but they were a bit more different than my case is.

Answer 1

You should equalize to same format d.m.Y and you can do this with strtotime and date function:

$post_date = date("d.m.Y", strtotime($_POST["date"]));

if($post_date == date("d.m.Y")){ 
  $headers.="Bcc: [e-mail]
";
}

I changed date to $post_date for more clear. I'll try to explain difference with outputs

echo $_POST["date"]; // lets say: 8.7.2013

echo date("d.m.Y"); // 09.09.2013 > it's current day

strtotime($_POST["date"]); // 1373230800 > it's given date with unix time

$post_date = date("d.m.Y", strtotime($_POST["date"])); // 08.07.2013 > it's given date as right format

If you use date function without param, it returns as current date.

Otherwise if you use with param like date('d.m.Y', strtotime('given_date'));, it returns as given date.

$post_date = date("d.m.Y", strtotime($_POST["date"]));

At first, we converted your date string to unix with strtotime then equalized and converted format that you used in if clause.