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Web 2016


Maladin

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[Message d'introduction]Version des files : Web 2016Domaine (Base de données, Core, etc.) : site webVotre niveau (débutant, intermédiaire, avancé)intermédiaireDescription du problème : warning mysql_fetch_object() supplied argument is not a valid mysql result resource

    Duyurular	<?PHPif(isset($_GET['id']) && checkInt($_GET['id'])){  $sqlNews = "SELECT * FROM ".SQL_HP_DB.".news WHERE anzeigen>0 AND id='".mysql_real_escape_string($_GET['id'])."' LIMIT 1";  $qryNews = mysql_query($sqlNews,$sqlHp);  if(mysql_num_rows($qryNews)>0)  {    while($getNews = mysql_fetch_object($qryNews))    {      echo'';      echo'['.$newsKategorien[$getNews->kategorie].'] '.$getNews->titel.'';      echo'
'.x_nl2br($getNews->inhalt).'';      if($newsKategorien[$getNews->kategorie]=='Event')      {        echo'
Düzenleme tarihi '.getDatum($getNews->datum).' saat '.getZeit($getNews->datum).' ';      }      else      {        echo'
Eklenme tarihi '.getDatum($getNews->datum).' saat '.getZeit($getNews->datum).' ';      }      echo'';    }  }  else   {    echo'
Geçerli ID bulunamadı.';  }}else{  $sqlNews = "SELECT * FROM ".SQL_HP_DB.".news WHERE anzeigen>0 AND kategorie!=2 ORDER BY datum DESC LIMIT 10";  $qryNews = mysql_query($sqlNews,$sqlHp);    while($getNews = mysql_fetch_object($qryNews))  {    $newsText = (strlen($getNews->inhalt)>400) ? substr($getNews->inhalt,0,strpos($getNews->inhalt,' ',400)).'... Devam..' : $getNews->inhalt;    echo'';    echo'['.$newsKategorien[$getNews->kategorie].'] '.$getNews->titel.'';    echo'
'.x_nl2br($newsText).'';    echo'
Yayınlanma tarihi '.getDatum($getNews->datum).' saat '.getZeit($getNews->datum).' ';    echo'';  }}?>    	

je ne vois pas de ou sa vient svp help En postant ma demande d'aide, j'affirme avoir lu et accepté le règlement des demandes d'aides.[Message d'introduction]

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