Mysql To_days

[Arc]

This is not returning anything. Am I Doing this right? I am trying to return the number of days between these two dates. "LastChanged' Is a field in the database containing a Date.

$UserID=$_SESSION["SESSION_USERID"];
$Today=date("Y-m-d");
$Sql="SELECT ChangePass, TO_DAYS(NOW())-TO_DAYS('LastChanged') AS TheDays   FROM tblUsers WHERE UserID ='$UserID'";
$Result=mysql_query($Sql) or die (mysql_error);
$ResultArray=mysql_fetch_array($Result);
Print $ResultArray['TheDays'];
if ($ResultArray['ChangePass']=='YES'){
  Print $ResultArray['TheDays'];
}

[LordShryku]

I think what you really want to use is datediff(though only supported since 4.1.1). Simple PHP method around this though

$Sql         = "SELECT ChangePass, LastChanged
                  FROM tblUsers 
                 WHERE UserID ='$UserID'";
$Result      = mysql_query($Sql) or die (mysql_error); 
$ResultArray = mysql_fetch_array($Result);
$TheDays     = ((time() - strtotime($ResultArray['LastChanged']) * 86400);

[Arc]

Thanks, atleast that is not giving me an error. However, it is returing a very long number rather than days. I assume the number is in seconds? -93483853940055 is the number.

What's the easiest way to convert that to days?

Thanks!😃

[LordShryku]

Oops :eek:

Should be dividing by 86400, not multiplying 😃

[Arc]

hmm, when I tried divide I got the number "1084289835.8333".

$TheDays = (time() - strtotime($ResultArray['LastChanged']) / 86400);

[LordShryku]

It's all good in theory 😃
Add some parens

$TheDays = ((time() - strtotime($ResultArray['LastChanged'])) / 86400);

[Arc]

That doesn't work either. I went ahead and did it the long way.

$TheSeconds = (time() - strtotime($ResultArray['LastChanged']));
$TheMinutes=$TheSeconds/60;
$TheHours=$TheMinutes/60;
$TheDays=round($TheHours/24);

Thanks!😃

[Weedpacket]

Odd - that's equivalent - only with a round().