my 'login date function' gives error reports

andreea115

hello everyone

i built a function to be able to determine when a user last logged into the site. The function works perfectly. However, when it comes to the last part of the function an error report is produced .

its best if i show u the function.





    function logindate( $datelogin= NULL   ) {



//the fuction takes a timestamp of the date that a user last logged in, this was // taken from the database.

// i now get a unix timestamp of the current date and time .  


 $currentdate = time (); 



  //i simple substract the date of the users last login from the current date
//i then use an if test to determine whether the person logged in within the last //two days 
//if not i then  convert the unixtime stamp back to a readable date using 
//the  DATE function. 

$logindate  = (  $currentdate-    $datelogin    );

if (( "$logindate" >=  '00001')  &&   ( "$logindate" <=  '86400') )    {

	echo' Today';


}
elseif (( "$logindate" >=  '86401')  &&   ( "$logindate" <=  '172800') )  

{

	echo' Yesterday ';


}


else 
{


         $date   =  date("d M Y " , $datelogin );  

         return   $date  ;
}


}


//i am calling the function and giving it a UNIX timestamp: this date would normally come from the database 

 logindate('500835843'   ) ;

the problem is that when i try to convert the date back to a human readable date it produces the following error report

A non well formed numeric value encountered

the error report relates to the $datelogin variable that i placed within the

$date = date("d M Y " , $datelogin );

when i however substituted the '$datelogin variable' for an actual number it work perfectly , and no error report was produced. So the problem must be in the way i have used the variable.

thank you everyone for your kind help

warm regards

Andreea

bradgrafelman

Why not just store the datetime directly in the DB, e.g. "2011-11-14 13:36:00" ?

andreea115

Hi bradgrafelman

i understand your suggestion and i will obviously try that when i make a new database. but the problem now is that i already have a database with several hundred users. so, i think its might be best to contiue as i stated.

accordingly, i am not clear why i am getting error reports. do you know why?

warm regards

Andreea

NogDog

For debugging purposes, I'd suggest something like this, in order to see what value the function is actually working with:

function logindate( $datelogin= NULL   ) { 
   // DEBUG:
   echo "<pre>"; var_dump($datelogin); echo "</pre>";
   // rest of function...
}

At least then you'll know exactly what is being fed into the function, and that may help us isolate the likely source of the problem.

andreea115

hi NogDog,

i tried the var_dump function e.g

echo "<pre>";   var_dump($datelogin); echo "</pre>";

and the value that i received for the variable $datelogin was ;

string '04 ,10 ,10, 11,2,2011' (length=21)

but would should be in the variable is

500835843

do you have any idea why PHP is converting the figure to this strange format.

bradgrafelman

Can you show us the full code you used to test this? PHP definitely isn't converting the data on its own...

andreea115

hello bradgrafelman

you have asked to see the full code. i now enclose it





function logindate( $datelogin= NULL   ) {






echo "<pre>";   var_dump($datelogin); echo "</pre>";


 $currentdate = time (); 



$logindate  = (  $currentdate-    $datelogin    );

if (( "$logindate" >=  '00001')  &&   ( "$logindate" <=  '86400') )    {

	echo' Today';


}
elseif (( "$logindate" >=  '86401')  &&   ( "$logindate" <=  '172800') )  

{


	echo' Yesterday ';


}


elseif (( "$logindate" >=  '172801')  &&   ( "$logindate" <=  '259200') )  

{


	echo'  3 Days Ago';


}


elseif (( "$logindate" >=  '259201')  &&   ( "$logindate" <=  '345600') )  

{


	echo'  4 Days Ago';


}

else 
{
           $date   =  date("d M Y " , $datelogin );  

         echo "<pre>";   var_dump($datelogin); echo "</pre>";
}




}


//call the function
   logindate("500835843"    ) ;

bradgrafelman

First problem is that the code is insanely hard to read... W h y u s e s o m u c h w h i t e s p a c e ? ? ?!

Next, why are you performing numerical comparisons against strings? Is "watermelon" less than or equal to "marshmallow" ? Is "Lorem ipsum" greater than "Hello world!" ?

andreea115

its not meant to be a string. its a unix timestamp. i.e
if (( "$logindate" >= '00001') && ( "$logindate" <= '86400') )

the above is a 24 hour time stamp. also, the variable $logindate is also a timestamp. its the diffrence between the current time and the last time the user logged in.

forgive me, but i am still new to PHP, why do you say its a string?

Weedpacket

andreea115 wrote:
forgive me, but i am still new to PHP, why do you say its a string?

Because they're all in quotation marks. That makes them strings.

   logindate("500835843"    ) ;

Is that really the sort of number you expect? It refers to a date in 1985. For current logins I would expect timestamps closer to 1321290243.

andreea115

Thank you to everyone for your kind help

i have now resolved the problem. the function now works perfectly.

warm regards

Andreea