小弟碰到一难题 求大神帮忙

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Release: 2016-06-13 12:32:15
Original
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小弟遇到一难题 求大神帮忙
 


    
              $rs=mysql_query("select * from title");
$row=mysql_fetch_array($rs);
$num=mysql_num_rows($rs);
//echo $num;
for($i=1;$i {
$rs=mysql_query("select * from title where parent_id=$i");
$row=mysql_fetch_array($rs);
//print_r($row)
?>
    
         小弟碰到一难题 求大神帮忙 " width="160px">
        

        
         $sql="select * from index_kind  where parent_id=$i";
$rs=mysql_query($sql);

while($row=mysql_fetch_array($rs))
{

?>
        ')" onMouseOut="omout(this)" id="">
      
        

        
         }
 ?>
        
        

        
        

         
" class="spfl_right">
       
        

   
    

    



(这是ajax的处理)

require_once 'conn.php';
$child_id=$_GET["id"];
$rs=mysql_query("select * from index_kind where child_id='$child_id'");
$row=mysql_fetch_array($rs);
$nums=mysql_num_rows($rs);
for($i=1;$i {
echo $row["good_img"];
}
?>


(这里是AJAX)
function omo(obj,str)
{
//alert(str);
obj.style.borderBottom="1px solid #e4393c";
//alert(obj.id);
 var xmlhttp;
     if(window.XMLHttpRequest)
     {
      xmlhttp=new XMLHttpRequest();
     }
     else
     {
         xmlhttp=new ActiveXObject("Microsoft.XMLHttp");
     }
  xmlhttp.open("get","chuli.php?id="+obj.id,true);
  xmlhttp.onreadystatechange=function(){
            //alert(xmlhttp.readyState)
              if(xmlhttp.readyState==4)
              {
                 alert(xmlhttp.responseText)
 
 document.getElementById(str).innerHTML=" 小弟碰到一难题 求大神帮忙 ";
 
              }
         }
     xmlhttp.send(null);
}

这是我测试用的数据库



然后当我移到这个划红线的地方是弹出的是图片的地址




我想让图片显示在下方的黑框里  但是弄不出来啊  !!! 求大神帮帮忙解决下问题!!!
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