AkMaL
  1. function newscom($var)
  2. {
  3. return mysql_result(mysql_query("SELECT COUNT(id) FROM news_comm WHERE user = '$var'"), 0);
  4. }
  5.  
  6. function status($var)
  7. {
  8. $ChEcKeD = newscom($var);
  9. if ($ChEcKeD == 0 || $ChEcKeD <= 15)
  10. {
  11. $status = '<span style="color:#0000FF;">Yangilardan</span>';
  12. $stars = $ChEcKeD;
  13. }
  14. if ($ChEcKeD >= 16 || $ChEcKeD <= 30)
  15. {
  16. $status = '<span style="color:#0000FF;">aktiv</span>';
  17. $stars = $ChEcKeD;
  18. }
  19. if ($ChEcKeD >= 31 || $ChEcKeD <= 50)
  20. {
  21. $status = '<span style="color:#0000FF;">faol</span>';
  22. $stars = $ChEcKeD;
  23. }
  24. elseif ($ChEcKeD >= 51)
  25. {
  26. $status = '<span style="color:#0000FF;">Sayt faxri</span>';
  27. $stars = $ChEcKeD;
  28. }
  29. else
  30. {
  31. $status = NULL;
  32. $stars = NULL;
  33. }
  34. return $status . (isset($stars) ? '<br/>' : '') . $stars . '<br/>';
  35. }


shuni qayerida xato bor 1-elseif dan o'tolmayapti
40 soniyadan keyin yozdi:
yangilikdagi izohlarga qarab unvon berishi kerak