MySQL的布爾給出，但結果都顯示

Question

，當我從MySQL顯示的數據我得到這樣的警告：

警告：mysql_fetch_assoc（）預計參數1是資源，在/ home/zh004600/www_root /公/文給定的boolean /atomia/find.php在線821

但我看到結果。我能做些什麼不會有這個警告？第821行與while循環「while（$ row1 = mysql_fetch_assoc（$ result1））{」

任何人都可以幫忙嗎？

echo '<table class="files" cellspacing="3" cellpadding="3">'; 

            $query = 

'SELECT * FROM files WHERE cz ='.$Podp; 
           $result = mysql_query($query); 
           $resultCount = mysql_num_rows($result) + 3; 

           for ($k=0;$k<$resultCount;$k+=4) { 
            $j= $k-4; 
             echo '<tr class="files">'; 
              if ($j == 0) { 
               $query1 = 'SELECT * FROM files WHERE cz ='.$Podp.' LIMIT 4'; 
              } else { 
               $query1 = 'SELECT * FROM files WHERE cz ='.$Podp.' LIMIT 4 OFFSET '.$j; 
              } 
               $result1 = mysql_query($query1); 
                while ($row1 = mysql_fetch_assoc($result1)) { 
                 $nazov = $row1['nazov']; 
                 $cesta = $row1['cesta'];        
                 echo '<th class="files"><a href="'.$cesta.'" download><img src="img/icons/pdf.png" width="45"></a><br>'.$nazov.'</th>'; 

                } 
             echo '</tr>'; 
           } 
         echo '</table>'; 

Answer 1

只需添加這個檢查

if($result1){ 

這樣的：

$result1 = mysql_query($query1); 
    if($result1){ // <-- THIS CHECK 
     while ($row1 = mysql_fetch_assoc($result1)) { 
      $nazov = $row1['nazov']; 
      $cesta = $row1['cesta'];        
      echo '<th class="files"><a href="'.$cesta.'" download><img src="img/icons/pdf.png" width="45"></a><br>'.$nazov.'</th>'; 

     } 
    }