PHP Date Help

[TomG1989]

Hi Guys, been stuck with thise for a few days now..

I've asked a few forums but still can't get it to work.

I have a signup script, that adds to a text field on a SQL database, this is working and add's the current date and time of the users signup.

On a page that displays there information i wan't this to state

"Member for: years, days, Minutes, Seconds."

I really can't seem to get this working, would someone be able to advise how i can do this?

The mysql field is $datesignup and need to find out how to write this!!

Please help! =]

[paulnaj]

Look up the date function in the manual. This formats dates/times as you require

[TomG1989]

Hi i have tried to look at articles for the last 2 days and stil can't get anything to work!

[paulnaj]

The link is not an article, but exact instructions on how to format dates/times.

In order to use any script, you need to get used to looking in the manual.

[NogDog]

This does not appear to be something for [man]date/man, as it's not an issue of formatting a given date, but in getting the difference between two date/time values and expressing them in those units. You might want to take a look at this function I wrote some time ago and see if you can incorporate it.

[TomG1989]

i have changed the signup field to now register as '1280907707' for example. instead of the actual date.

I can display this as the date on my page... still cannot integrate anything to display it as Days/Minutes/Seconds ....

ive tried all sorts! All i wan't is it to state Memeber for : days, minutes, **seconds

[johanafm]

Did you take a look at the link in NogDog's last post? Did you try the code he provided you with?

[Crowly]

Something like this should work, using NogDogs function with months and weeks removed, and i presume mysql

function timeDiff($t1, $t2)
{
   if($t1 > $t2)
   {
      $time1 = $t2;
      $time2 = $t1;
   }
   else
   {
      $time1 = $t1;
      $time2 = $t2;
   }
   $diff = array(
      'years' => 0,
      'months' => 0,
      'weeks' => 0,
      'days' => 0,
      'hours' => 0,
      'minutes' => 0,
      'seconds' =>0
   );
  //foreach(array('years','months','weeks','days','hours','minutes','seconds')
   foreach(array('years','days','hours','minutes','seconds')
         as $unit)
   {
      while(TRUE)
      {
         $next = strtotime("+1 $unit", $time1);
         if($next < $time2)
         {
            $time1 = $next;
            $diff[$unit]++;
         }
         else
         {
            break;
         }
      }
   }
   return($diff);
}

$sql="SELECT DATE_FORMAT(datesignup,'%Y-%m-%d %T'), DATE_FORMAT(NOW(),'%Y-%m-%d %T')
FROM table
WHERE something....";

$row=mysql_fetch_row($mysql_query($sql));

$start = strtotime($row[0]);
$end = strtotime($row[1]);
$diff = timeDiff($start, $end);

echo "Member for: ".$diff['years']." years, ".$diff['days']." days, ".$diff['minutes']." Minutes, ".$diff['seconds']." Seconds.";

[TomG1989]

Parse error: syntax error, unexpected ')'

somewhere in that

[TomG1989]

Fatal error: Function name must be a string

[Crowly]

Can't see anything wrong with that script, it works here. Of course the sql isn't complete, but that should be pretty obvious. Also this isn't a complete script, you need to add code to connect to the database and so forth. So i suggest you post your complete script if you can't get this to work.

[vividona]

I don't know if I understand you
but you can try this:

1. Make time in mysql as varchar.

2. In the data insertion process you have to insert time as follow

   $time = time();     
   
   insert into table (time) values('".$time."')

3. when you fetch your query you have to do something like this

   while($row = mysql_fetch_array($query)){
   	$registerTime = date("j, n, Y g:i a", $row['time']);
   	echo $registerTime;
   }

[TomG1989]

$result = mysql_query("select id, name, datesignup from members where id=$_GET[id]");

<?function timeDiff($t1, $t2)
{
if($t1 > $t2)
{
$time1 = $t2;
$time2 = $t1;
}
else
{
$time1 = $t1;
$time2 = $t2;
}
$diff = array(
'years' => 0,
'months' => 0,
'weeks' => 0,
'days' => 0,
'hours' => 0,
'minutes' => 0,
'seconds' =>0
);
//foreach(array('years','months','weeks','days','hours','minutes','seconds')
foreach(array('years','days','hours','minutes','seconds')
as $unit)
{
while(TRUE)
{
$next = strtotime("+1 $unit", $time1);
if($next < $time2)
{
$time1 = $next;
$diff[$unit]++;
}
else
{
break;
}
}
}
return($diff);
}

$sql="SELECT DATE_FORMAT(datesignup,'%Y-%m-%d %T'), DATE_FORMAT(NOW(),'%Y-%m-%d %T') FROM table members WHERE id=$_GET[id]";

$row=mysql_fetch_row($mysql_query($sql));

$start = strtotime($row[0]);
$end = strtotime($row[1]);
$diff = timeDiff($start, $end);

echo "Member for: ".$diff['years']." years, ".$diff['days']." days, ".$diff['minutes']." Minutes, ".$diff['seconds']." Seconds."?>

My SQL Field is - int(12) and gets entered in when a user signs up with a timestamp.

[Crowly]

Please put your code/script in php brackets (without the )

[*php][*/php]

TomG1989;10960058 wrote:

$result = mysql_query("select id, name, datesignup from members where id=$_GET[id]");

Don't think this part is needed, comment it out for the time being.

$sql="SELECT DATE_FORMAT(datesignup,'%Y-%m-%d %T'), DATE_FORMAT(NOW(),'%Y-%m-%d %T') FROM table members WHERE id=$_GET[id]"; 

This is wrong, its not FROM table members, just FROM members. Also i forgot to mention, that including something like $_GET['id'] (user submited data) directly into a query without validating it first is a security risk, and i made the same mistake myself in the previous example :o. So make sure that you know that $_GET['id'] is a number.

Try this

function timeDiff($t1, $t2) 
{ 
   if($t1 > $t2) 
   { 
      $time1 = $t2; 
      $time2 = $t1; 
   } 
   else 
   { 
      $time1 = $t1; 
      $time2 = $t2; 
   } 
   $diff = array( 
      'years' => 0, 
      'months' => 0, 
      'weeks' => 0, 
      'days' => 0, 
      'hours' => 0, 
      'minutes' => 0, 
      'seconds' =>0 
   ); 
  //foreach(array('years','months','weeks','days','hours','minutes','seconds') 
   foreach(array('years','days','hours','minutes','seconds') 
         as $unit) 
   { 
      while(TRUE) 
      { 
         $next = strtotime("+1 $unit", $time1); 
         if($next < $time2) 
         { 
            $time1 = $next; 
            $diff[$unit]++; 
         } 
         else 
         { 
            break; 
         } 
      } 
   } 
   return($diff); 
} 

// it might not be needed to use DATE_FORMAT on NOW() since i think i format it to NOW()'s default output
$sql=sprintf("SELECT DATE_FORMAT(datesignup,'%%Y-%%m-%%d %%T') datesignup, 
DATE_FORMAT(NOW(),'%%Y-%%m-%%d %%T') cur_time, id, name
FROM members 
WHERE id=%d",
mysql_real_escape_string($_GET['id'])); 

$row=mysql_fetch_row($mysql_query($sql)); 

$start = strtotime($row[0]); 
$end = strtotime($row[1]); 
$diff = timeDiff($start, $end); 

echo "Member for: ".$diff['years']." years, ".$diff['days']." days, ".$diff['minutes']." Minutes, ".$diff['seconds']." Seconds.";