Ajax调用后图像未显示在页面上

Question

我想知道您是否能再次帮助我...

我正在使用ajax调用php页面，以将clientwidth值传递给它，以便可以适当调整图像大小。我从哪里进行ajax调用的位置在调用页面的“主” div内，但是未显示图像。你能帮忙吗？

我的ajax呼叫：

 <script>
     // Get the HTTP Object
function getHTTPObject(){
   if (window.ActiveXObject) 
       return new ActiveXObject("Microsoft.XMLHTTP");
   else if (window.XMLHttpRequest) 
       return new XMLHttpRequest();
   else {
      alert("Your browser does not support AJAX.");
      return null;
   }
}
w=document.documentElement.clientWidth;
    // Implement business logic    
    httpObject = getHTTPObject();
    if (httpObject != null) {
        httpObject.open("GET", "imagedisplay.php?width=" + w + "&id=" + <?php print $row['id'] ?> +"&imageno="+<?php print $imageno ?>, true);
        httpObject.send(null);

    }
var httpObject = null;
</script>  

...以及相关的php文件...

<?php
 include "connection.php";
    $id=$_GET["id"];
    $w=$_GET["width"];
    $imageno=$_GET["imageno"];
    $query = "Select * from courses where id = '$id'"; 
    $result = mysqli_query($connection,$query) or die ("Error in query: $query. ".mysqli_error($connection));
    while($row = mysqli_fetch_array($result)) {        
    $table="<table><tr>";
        switch ($imageno) {
        case 1:
        $photowidth=$w;
        break;
        case 2:
        $photowidth=$w/2;
        break;
        case 3:
        $photowidth=$w/3;
        break;
        case 4:
        $photowidth=$w/2;
        break;
        case 5:
        $photowidth=$w/3;
        break;
        case 6:
        $photowidth=$w/3;
        break;
    }
    if ($row['iamge1']!=NULL) {
        $table .="<td><img src=\"../".$row['image1']."\" width=\"".$photowidth."\" border=\"0\"></td>";
        }
        if ($row['image2']!=NULL) {
        $table .="<td><img src=\"../".$row['image2']."\" width=\"".$photowidth."\" border=\"0\"></td>";
    }
        if ($row['image3']!=NULL) {
        $table .="<td><img src=\"../".$row['image3']."\" width=\"".$photowidth."\" border=\"0\"></td>";
        }
        if ($row['image4']!=NULL) {
        $table .="<td><img src=\"../".$row['image4']."\" width=\"".$photowidth."\" border=\"0\"></td>";
        }
        if ($row['image5']!=NULL) {
        $table .="<td><img src=\"../".$row['image5']."\" width=\"".$photowidth."\" border=\"0\"></td>";
        }
        if ($row['image6']!=NULL) {
        $table .="<td><img src=\"../".$row['image6']."\" width=\"".$photowidth."\" border=\"0\"></td>"; 
        }
    }
    $table.="</tr><table>";
    print $table;
    ?>

我知道正在调用php页面，因为当我在其中引入一些错误时，它在apache日志中报告了。我还手动将键/值对输入到文件中，然后图像出现在页面上。提前...