錯誤警告：mysql_fetch_assoc（）期望參數1是資源，布爾值給出

Question

我收到此消息時，我的數據庫表沒有記錄 所以如何解決它。

$sql2 = mysql_query("SELECT ID,NAME from login where TYPE='AdPublish' ORDER BY ID ASC $limit") or mysql_error(); 

if(!$sql2) 
{ 
    die("invalid query".mysql_error()); 
} 
else 
{ 
    while($data=mysql_fetch_array($sql2)) 
    { 
     $rv1=mysql_query("select LINK from url where ID='".$data['ID']."' AND STATUS='Pending'") or mysql_error(); 
     $count1=mysql_num_rows($rv1); 
     echo '<a href="cidmatter.php?cid='.$data['ID'].'" target="_blank">'; 
     echo "<table>"; 
     echo '<tr><td><h4><b>ID: ' .$data['ID'].'</b></H4></td></tr><br>'; 
     echo '<br><tr><td><b>Name: ' .$data['NAME'].'</b>('.$count1.')</td></tr></table></a>'; 
     echo '<hr>'; 
    } 
} 

Answer 1

使用這樣的事情

$sql2 = mysql_query("SELECT ID,NAME from login where TYPE='AdPublish' ORDER BY ID ASC $limit") or die("Error in query"); 

    while($data=mysql_fetch_array($sql2)){ 
     $rv1=mysql_query("select LINK from url where ID='".$data['ID']."' AND STATUS='Pending'") or die("Error in query"); 
     $count1=mysql_num_rows($rv1); 
     echo '<a href="cidmatter.php?cid='.$data['ID'].'" target="_blank">'; 
     echo "<table>"; 
     echo '<tr><td><h4><b>ID: ' .$data['ID'].'</b></H4></td></tr><br>'; 
     echo '<br><tr><td><b>Name: ' .$data['NAME'].'</b>('.$count1.')</td></tr></table></a>'; 
     echo '<hr>'; 

}