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supplied argument is not a valid MySQL result

开发者 https://www.devze.com 2022-12-27 12:30 出处:网络
I have written a function: function selectWithPaging($where){ $pg = (int) (!isset($_GET[\"pg\"]) ? 1 : $_GET[\"pg\"]);

I have written a function:

function selectWithPaging($where){

 $pg = (int) (!isset($_GET["pg"]) ? 1 : $_GET["pg"]);
 $pg = ($pg == 0 ? 1 : $pg);
 $perpage = 10;//limit in each page
 $startpoint = ($pg * $perpage) - $perpage;
 $result = mysql_query("SELECT * FROM $where ORDER BY id ASC LIMIT $startpoint,$perpage");
 return $result;
               }

but when inserting in this function :

function categories() { 

 selectWithPaging('category') 
 $text .='<h2 class="mainH">Categories</h2>';
 $text .= '<table><tr class="cn"><td>ID</td><td class="name">Category</td>  <td>Durum</td>'; 
 while ($row = mysql_fetch_array($result)) {
 $home    = $row['home']; 
 $publish = $row['published'];
 $ID = $row['id'];
 $src = '<img src="'.ADMIN_IMG.'homec.png"开发者_如何学C>';
                      -------------
          }

there is this error: supplied argument is not a valid MySQL result

What is wrong in my first function?


This may be a typo, but your first function returns a result that isn't being caught by any variable in your second function.

change:

selectWithPaging('category')  

to:

$result = selectWithPaging('category');  

and give it a try.


Your query isn't running. It's either because you haven't passed the $dblink arg to mysql_query, or because your SQL syntax has an error. This should tell you what's happening.

$dblink=mysql_connect('localhost', 'mysql_user', 'mysql_password');
try{
  $query="SELECT * FROM $where ORDER BY id ASC LIMIT $startpoint, $perpage";
  $result = mysql_query($query,$dblink) or throw new Exception(mysql_error($dblink));
  return $result; 
} catch(Exception $e){
    echo $e->getMessage();
} 
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