PHP-AJAX 多表单发布

Question

在我的 PHP 应用程序的一个模块中，有多个表单，由服务器端生成的唯一 ID 和名称为“formToProcess1、formToProcess2、...、fromToProcessN”。但是，有两个 ajax 函数处理每个表单的请求，它们都以字符串形式获取表单名称作为参数，如下所示：

 function ajaxReject(formToProcess)
    {
     var ajaxRequest;
        //document.getElementById('divUploadResultAjax').style.display = 'none';



        try
        {
          ajaxRequest = new XMLHttpRequest();
        }
        catch (e)
        {
          try
          {
             ajaxRequest = new ActiveXObject('Msxml2.XMLHTTP');
          }
          catch (e) 
          {
             try
             {
                ajaxRequest = new ActiveXObject('Microsoft.XMLHTTP');
              }
              catch (e)
              {
                alert('Something is wrong with your browser, AJAX not working!');
               return false;
             }
          }
        }

        ajaxRequest.onreadystatechange = function()
        {
          if(ajaxRequest.readyState == 4)
          {
             var ajaxDisplay = document.getElementById('divUploadResultAjax');
             ajaxDisplay.innerHTML = ajaxRequest.responseText;
             //document.getElementById('loadingAjaxIcon').style.display = 'none';
             document.getElementById('divUploadResultAjax').style.display = 'block';
          }
        }

        var updName         = document.forms['formToProcess'].getElementById('userName').value;

                      //i'll get other elements values for query string to send if i can get this one first :)




        var queryString = '?name=' + updName + '&lastname=' + updLastname + '&bio=' + updBio + '&country=' + updCountry + '&cams=' + updCams + '&fb=' + updFacebook + '&twitter=' + updTwitter + '&processRequest=' + RequestProcess;

        alert (queryString);
        return;


        ajaxRequest.open('GET', 'dashboardUpdateProfile.php' + queryString, true);
        ajaxRequest.send(null); 
                        }



#Every form in the php page is generated like following:

        echo '

                                        <div id="divUploadResultAjax">
                                            <form name="profileUpdater'. $formCounter .'">
                                                <input type="hidden" id="uluid" name="uluid" value="'.$UL_UploadID.'" />
                                                <table class="tg" style="width: 100%">
                                                    <tr>
                                                        <td style="font-weight: bold; width: 250px; color: #000000;" colspan="3">
                                                            <h3>Upload ID #'.$UL_UploadID.' </h3> (Uploaded On '.$NewCreateDate.')
                                                        </td>
    ...
    ...
    bla bla bla
    ...
    ...

    ----> this is where i call functions in each form.
    <input type="button" class="btnRegister" name="updateProfile"     onclick="ajaxReject(\'profileUpdater'. $formCounter .'\')" value="Reject" />
             <input type="button" class="btnRegister" name="updateProfile" onclick="ajaxApprove()"     value="Approve" />

我不太确定出了什么问题。当我通过单击按钮调用该函数时，它可以完美运行，直到 if(ajaxRequest.readyState == 4) 结束，之后，当我尝试获取 var updName = document.forms['formToProcess'].getElementById( '用户名').value;功能崩溃。我做错了什么？

谢谢。

Answer 1

你不能用var updName = document.getElementById('userName').value;之类的东西吗？

我没有在表单中找到用户名元素？它在你的实际代码中吗？