如果php中的语句使用日期

I want to write an IF statement based on two dates.
I have a MySQL database (of which I have no control over) for tests and exams taken.
One of the fields is labelled ClientTime and outputs the date and time the test was taken. When queried the typical output would read as such Thu Aug 25 16:47:05 GMT+0100 2011

The following all happens in a while loop

I’ve managed to convert this MySQL date using strtotime() to a d/m/Y format so the above would read 25/08/2011 and I can display a table with all the correct dates.

I also have a php form that asks for the number of days that this test will expire after and this gets stored in a variable $IMonths (from the $_POST['IMonths'] passed from the form page) - the php code will be changed to Months once I have it working but I am using days for the purpose of testing – hence the variable passed being called $IMonths.

I have also created a $today = date("d/m/Y"); variable

I can successfully create a new date using the following command $Expiry = date('d/m/Y', strtotime($row['ClientTime'] . "+$IMonths days")); and can display all results with both the test date and the expiry date.

I then want to include an if-statement within my while-statement that says is ($Expiry>$today) but it’s not accurate – there are no errors and changing the number of days changes the results.

My test database has 5 exams in with the following dates.

Taken 17/08/2011
Taken 17/08/2011
Taken 22/08/2011
Taken 22/08/2011
Taken 24/08/2011

If I set the expiry days to 1, I get all results back as expected:

If I set it to 10 days I get two results

Taken 17/08/2011
Taken 17/08/2011

I don’t get any of the others even though their date will be less than today's date (07/09/2011 (as I write this)) but they do expire in the month of September.

If I set the expiry days to 15, I get one result that is one day in the future!

Taken 17/08/2011

I hope that makes sense. My code is as follows:

<?php

// Make The MySQL Server Connection
mysql_connect("localhost", "root", "") or die(mysql_error());

// Make The MySQL Database Connection
mysql_select_db("questions") or die(mysql_error());

$IQuizName = $_POST['IQuizName'];
$IMonths = $_POST['IMonths'];

$result = mysql_query("SELECT * FROM questions.resultsets 
                       WHERE QuizName='$IQuizName' 
                       AND PassFail='Pass'") or die (mysql_error());

$today = date("d/m/Y");

echo "<table border='1'>";
echo "<tr> <th align='left'>Candidate Name </th> <th align='left'>Pass / Fail</th> <th align='left'>Date last Taken</th> <th align='left'>Expires</th> <th align='left'>Today</th> </tr>";

  while($row = mysql_fetch_array($result)){

  //START of expiry calculation
  $Expiry = date('d/m/Y', strtotime($row['ClientTime'] . "+$IMonths days"));

  if($Expiry>$today){

    echo "<tr><td>"; 
    echo $row['Candidate'];
    echo "</td><td>";  

    echo $row['PassFail'];
    echo "</td><td>"; 

    //Date Conversion
    echo date('d/m/Y', strtotime($row['ClientTime']));
    echo "</td><td>"; 

    echo $Expiry;
    echo "</td><td>"; 

    echo $today;
    echo "</td><tr>"; 

  };
}

echo "</table>";
echo "<P>";
?>

For accurate comparing dates you should convert the dates with the strtotime function again.

Your issue is due to datatypes in PHP. When you call the date() function in PHP, you are actually formatting dates into strings. So when you do $Expiry = date(blahblah), $expiry is now a string. So your if statement is if (somestring > someotherstring). Normally you would not think to compare one block of text to see if it is greater than another block of text.

http://www.php.net/manual/en/function.date.php

One way to get around this would be the strtotime function. This returns integers that are based on UNIX timestamp for the date representation.

http://www.php.net/manual/en/function.strtotime.php

If you convert your string dates to strtotime, your if statement is now comparing if one number is greater than another. This should solve your problem :)

First: fix that SQL-injection hole.

I suggest querying the expiry dates in SQL, it will be much easier.

Here's example code

//escape those user variables !!
$IQuizName = mysql_real_escape_string($_POST['IQuizName']);
$IMonths = mysql_real_escape_string($_POST['IMonths']);

$result = mysql_query("SELECT * FROM questions.resultsets 
                   WHERE QuizName= '$IQuizName' 
                   AND PassFail= 'Pass' 
                   AND ClientTime < NOW() "); 

//don't use (or die(error)) in production code, (for now I'll let it slide).
if (!$result) then { die (mysql_error()); }
while ($row = mysql_fetch_row($result)) 
{
  echo "the table"
}

In PHP there is no standard date data type. What you have is strings, and when you compare them, they get compared as strings, not as dates, thats why you get results you dont expect. How it normally is done in PHP is to format your date string by ordering from most significant bit to least bit, like

YYYY-MM-DD HH:MM:SS

If you do that, string comparison will work as expected. You can format your dates easily using:

$string = date('Y-m-d H:i:s', $year, $month, $day, $hour, $minute, $second);