How to debug my PHP validation function? - php

Question

I am creating a web form for my work which is being validated using PHP. However, when I test the page I keep getting all of my error messages returned without the form being submitted properly when valid information is inputted. The following is a small section of the code (including the HTML sections).
<?php
$date =""
$dateerror = ""
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["date"])) {
$dateerror = "Date is required";
} else {
$date = test_input($_POST["date"]);
$array = explode("/", $date);
$day = $array[1];
$month = $array[0];
$year = $array[2];
if (!checkdate($month, $day, $year)) {
$dateerror = "Date mustbe in M/D/Y format";
} else {
date_default_timezone_set("America/Anchorage");
$today = strtotime("now");
if (strtotime($date)>=$today) {
$date = test_input($_POST["date"]);
} else {
$dateerror = "Date is before present day";
}
}
}
<input type="text" size="9" name="date" id="date" required title="Please enter current date"><?php echo $dateerror; ?><br>
Again, the PHP code just returns "Date is before present day" even when the date is the current date.

Answer 1

If you want to validate a date in PHP, the best way to do it is to use the DateTime class, and specifically the createFromFormat method.
This call will create a DateTime object set to the specified date in the given format, or false if it was an invalid date.
So for example:
<?php
$input = "05/08/2015";
$test = DateTime::createFromFormat('d/m/Y', $input);
if (!$test) {
print "You entered an invalid date";
die;
}
$now = new DateTime();
if ($test < $now) {
print "Date is before present.";
die;
}
?>
Simple as that. There's no need for regex, or for exploding the input, etc; just a single simple test. And you can also then use the $test variable to process the date as well once you've determined that it's valid, since it's a standard DateTime object.
[EDIT] I've added a bit in the code to deal with using the DateTime class to handle date comparisons, to give the 'before present' error.
The important point here is that if you have a DateTime object, you need to compare it with another DateTime object; the older strtotime() produces a different type of date resource to DateTime, and you can't use them together (at least not without converting between them all the time).

Answer 2

The solution: use date("M/D/Y"):
$today = strtotime(date("M/D/Y")); // 1432958400
$date = strtotime($_POST["date"]); // user input. 05-30-2015 will yield 1432958400
// the rest of your logic here
Here's the code specific solution:
<?php
$date =""
$dateerror = ""
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["date"])) {
$dateerror = "Date is required";
} else {
$date = test_input($_POST["date"]);
$array = explode("/", $date);
$day = $array[1];
$month = $array[0];
$year = $array[2];
if (!checkdate($month, $day, $year)) {
$dateerror = "Date mustbe in M/D/Y format";
} else {
date_default_timezone_set("America/Anchorage");
$today = strtotime(date("M/D/Y"));
if (strtotime($date)>=$today) {
$date = test_input($_POST["date"]);
} else {
$dateerror = "Date is before present day";
}
}
}
<input type="text" size="9" name="date" id="date" required title="Please enter current date"><?php echo $dateerror; ?><br>