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长期聆听者，第一次来电者 :) 需要帮助 - 我的 Java Script 知识有限，我已经在这个问题上停留了一段时间。

我有一个实现 AJAX 方法的经典 ASP 脚本。对我来说挑战是我需要能够告诉 Java 脚本哪个 DIV ID 将 Ajax 调用返回的 html 加载到其中。

目标 DIV 可能因应用程序正在执行的操作而异。我的想法是将 DIV 名称作为参数传递给 Java Script。

这是我到目前为止所拥有的：

<div id="AjaxResponseDiv">Content displayed before the Ajax call is made</div>
<a href="javascript:CallAjaxPage('/dev/ajax/somePage.asp?par=1','AjaxResponseDiv');">Get Message From Server</a>

Java 脚本如下：

function XHConn()
{
  var xmlhttp, bComplete = false;
  try { xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); }
  catch (e) { try { xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); }
  catch (e) { try { xmlhttp = new XMLHttpRequest(); }
  catch (e) { xmlhttp = false; }}}
  if (!xmlhttp) return null;
  this.connect = function(sURL, sMethod, sVars, fnDone)
  {
    if (!xmlhttp) return false;
    bComplete = false;
    sMethod = sMethod.toUpperCase();
    try {
      if (sMethod == "GET")
      {
        xmlhttp.open(sMethod, sURL+"?"+sVars, true);
        sVars = "";
      }
      else
      {
        xmlhttp.open(sMethod, sURL, true);
        xmlhttp.setRequestHeader("Method", "POST "+sURL+" HTTP/1.1");
        xmlhttp.setRequestHeader("Content-Type",
          "application/x-www-form-urlencoded");
      }
      xmlhttp.onreadystatechange = function(){
        if (xmlhttp.readyState == 4 && !bComplete)
        {
          bComplete = true;
          fnDone(xmlhttp);
        }};
      xmlhttp.send(sVars);
    }
    catch(z) { return false; }
    return true;
  };
  return this;
}
var doAJAXCall = function (PageURL, ReqType, PostStr, FunctionName, AjaxResponseDiv) {

    // create the new object for doing the XMLHTTP Request
    var myConn = new XHConn();

    // check if the browser supports it
    if (myConn) {

        // XMLHTTPRequest is supported by the browser, continue with the request
        myConn.connect('' + PageURL + '', '' + ReqType + '', '' + PostStr + '', FunctionName);    
    } 
    else {
        // Not support by this browser, alert the user
        alert("XMLHTTP not available. Try a newer/better browser, this application will not work!");   
    }
}

// launched from button click 
var CallAjaxPage = function (URL, AjaxResponseDiv) {

    // build up the post string when passing variables to the server side page
    var PostStr = "";

    // use the generic function to make the request
    //alert(showMessageResponse)

    showMessageResponse = showMessageResponse

    doAJAXCall(URL, 'POST', '', showMessageResponse, AjaxResponseDiv);
}

// The function for handling the response from the server

var showMessageResponse = function (oXML) { 

    // get the response text, into a variable
    var response = oXML.responseText;

    // update the Div to show the result from the server
    document.getElementById("AjaxResponseDiv").innerHTML = response;
};

</script>

</code>

So essentially the question is - how can I change "AjaxResponseDiv" in line: 

document.getElementById("AjaxResponseDiv").innerHTML = response;

To a variable AjaxResponseDiv which I am catching here:

    // launched from button click <br/>
var CallAjaxPage = function (URL, AjaxResponseDiv)

4

1 回答 1

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The callback function should be a closure within the CallAjaxPage function.

var CallAjaxPage = function (URL, AjaxResponseDiv) {

    // build up the post string when passing variables to the server side page
    var PostStr = "";

    // use the generic function to make the request
    //alert(showMessageResponse)

    function showMessageResponse (oXML) {
        // get the response text, into a variable
        var response = oXML.responseText;
        // update the Div to show the result from the server
        document.getElementById(AjaxResponseDiv).innerHTML = response;
    }

    doAJAXCall(URL, 'POST', '', showMessageResponse);
}

doAJAXCall doesn't need the AjaxResponseDiv argument.

于 2013-05-24T02:58:23.430 回答