AJAXを介してページを読み込む

Question

<form> 
     <table> 
      <tr> 
       <td> 
        <input type="tex" id="arheading" name="arheading" value="Article Heading" onfocus="if (this.value == 'Article Heading') {this.value = '';}" onblur="if (this.value == '') {this.value = 'Article Heading';}" /> 
      </td> 
       <td> 
        <select id="plaetype" name="placetype"> 
         <option selected="Place Type">Place Type</option> 
         <?php 
         include 'ini/configs_adm.php'; 
         $result_adm = mysql_query("SELECT * FROM places"); 
         while($row_adm=mysql_fetch_array($result_adm)) 
         { 
          echo '<option value="'.$row_adm['places'].'">'.$row_adm['places'].'</option>'; 
         } 
         ?> 
        </select> 
       </td> 
       <td> 
        <select id="country" name="country"> 
        <?php 
         include 'ini/configs_adm.php'; 

         $result_adm = mysql_query("SELECT * FROM countries"); 
         while($row_adm=mysql_fetch_array($result_adm)) 
         { 
         echo '<option value="'.$row_adm['country'].'">'.$row_adm['country'].'</option>'; 
         } 
        ?> 
        </select>  
       </td> 
      </tr> 
     </table>  
    </form> 

上記は、AJAXをリフレッシュまたはナビゲーションせずにインデックスページにロードしたいコードです。

と私はしようとしたAJAXコード.........

var loadedobjects = "" 
var rootdomain = "http://" + window.location.hostname 

function ajaxpage(url, containerid) { 
    var page_request = false 
    if (window.XMLHttpRequest) // if Mozilla, Safari etc 
     page_request = new XMLHttpRequest() 
    else if (window.ActiveXObject) { // if IE 
     try { 
      page_request = new ActiveXObject("Msxml2.XMLHTTP") 
     } 
     catch (e) { 
      try { 
       page_request = new ActiveXObject("Microsoft.XMLHTTP") 
      } 
      catch (e) { 
      } 
     } 
    } 
    else 
     return false 
    page_request.onreadystatechange = function() { 
     loadpage(page_request, containerid) 
    } 
    page_request.open('GET', url, true) 
    page_request.send(null) 
} 

function loadpage(page_request, containerid) { 
    if (page_request.readyState == 4 && (page_request.status == 200 || window.location.href.indexOf("http") == -1)) 
     document.getElementById(containerid).innerHTML = page_request.responseText 
} 

function loadobjs() { 
    if (!document.getElementById) 
     return 
    for (i = 0; i < arguments.length; i++) { 
     var file = arguments[i] 
     var fileref = "" 
     if (loadedobjects.indexOf(file) == -1) { //Check to see if this object has not already been added to page before proceeding 
      if (file.indexOf(".js") != -1) { //If object is a js file 
       fileref = document.createElement('script') 
       fileref.setAttribute("type", "text/javascript"); 
       fileref.setAttribute("src", file); 
      } 
      else if (file.indexOf(".css") != -1) { //If object is a css file 
       fileref = document.createElement("link") 
       fileref.setAttribute("rel", "stylesheet"); 
       fileref.setAttribute("type", "text/css"); 
       fileref.setAttribute("href", file); 
      } 
     } 
     if (fileref != "") { 
      document.getElementsByTagName("head").item(0).appendChild(fileref) 
      loadedobjects += file + " " //Remember this object as being already added to page 
     } 
    } 
} 

とAJAX機能があるために呼び出すコード.....

<ul id="admin_menu"> 
    <?php 
    include 'ini/configs_adm.php'; 
    $result_adm = mysql_query("SELECT * FROM adm_tools"); 
    while ($row_adm = mysql_fetch_array($result_adm)) 
    { 
     echo '<li><a href="javascript:ajaxpage(\'' . $row_adm['adm_menu_lnk'] . '\', \'contentbox\');">' . $row_adm['adm_menu'] . '</a></li>'; 
    } 
    ?> 
</ul> 

私のAjaxコードがページを読み込んでいますが、PHPスクリプトがデータベースからデータを取得していません

Answer 1

extract（$ row）関数が見つからないようです。以下のようになります

：

while($row_adm=mysql_fetch_array($result_adm)) 
{ 
extract($row_adm); 
echo '<option value="'.$row_adm['places'].'">'.$row_adm['places'].'</option>'; 
}