2017-03-15 44 views
1

,当我从MySQL显示的数据我得到这样的警告:MySQL的布尔给出,但结果都显示

警告:mysql_fetch_assoc()预计参数1是资源,在/ home/zh004600/www_root /公/文给定的boolean /atomia/find.php在线821

但我看到结果。我能做些什么不会有这个警告?第821行与while循环“while($ row1 = mysql_fetch_assoc($ result1)){”

任何人都可以帮忙吗?

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>'; 

回答

2

只需添加这个检查

if($result1){ 

这样的:

$result1 = mysql_query($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>'; 

     } 
    } 
+0

谢谢,其作品 – Stanley01

+0

很高兴听到这个。当'mysql_fetch_assoc($ result1)'返回'false'而不是资源时,您发出了此警告。 –

+0

感谢您的解释 – Stanley01