counter

[IBEX]

	  $sql = "SELECT counter FROM video WHERE video_id = {$_GET['video_id']}";
  $result = mysql_query($sql) or die (mysql_error());
  $rows = mysql_fetch_array($result);

  $page_views = $rows['counter'];

  if (empty($page_views)) { 
	 $page_views = "1"; 
 mysql_query("INSERT INTO video (counter)VALUES('1') WHERE video_id = {$_GET['video_id']}");
  }



  $page_views++;
  $sql = "UPDATE video SET counter = $page_views WHERE video_id = {$_GET['video_id']}";
  $result = mysql_query($sql) or die (mysql_error());

 

Kas sheit ir nepareizi?.. man liekas ieksha nevis 1, bet 2..

[andrisp]

Viss taču ir pareizi - tu $page_views (kam vērtība ir 1) palielini par vienu, rezultātā sanāk 2.

[IBEX]

ka panakt lai rezultata butu 1?..

[andrisp]

Pieliec else. Tb, ja $page_views tomēr nav tukšs, tad updeito.

[codez]

vispār jau to var uzrakstīt vienā kverijā:

INSERT INTO video (video_id,counter) VALUES (444,1) ON DUPLICATE KEY UPDATE counter=counter+1;