如何使用jsp返回對AJAX的響應？

我是jsp的新手，所以我想知道如何返回AJAX響應。這是使用道場提交表單與AJAX代碼：如何使用jsp返回對AJAX的響應？

<script type="text/javascript"> 
     dojo.require("dojo.io.iframe"); 

     dojo.addOnLoad(function() { 
      upload = function() { 
      dojo.io.iframe.send({ 
       form : "fileUploader", 
       handleAs : "html", //response type from the server 
       url : "url to jsp",//just example, not the real url 
       load : function(response, ioArgs) { 
        console.log(response, ioArgs); 
        return response; 
       }, 
       error : function(response, ioArgs) { 
        console.log("error"); 
        console.log(response, ioArgs); 
        return response; 
       } 
      }); 
     }; 
     });

，這是JSP代碼返回響應片段：

r = "<p >You have successfully uploaded your video. </p>"; 
response.setContentType("text/html"); 
response.getWriter().write(r);

如果我只是用System.out.println（r），它在控制檯中打印結果，所以表單正在被提交。所以我認爲問題在於我回復的方式，或者我用Dojo處理它的方式。

在此先感謝！

來源

2013-06-20 ricardo28

這裏是簡單的AJAX和JSP（無jquery）來點擊一個按鈕，並獲得一個時間回到輸入框。看看「document.myForm.time.value = xmlhttp.responseText;」這一行，它在那裏得到響應並在jsp中設置結果。

<html> 
<head> 
<title>JSP and Servlet using AJAX</title> 
<script type="text/javascript"> 



function getXMLObject() //XML OBJECT 
{ 
    var xmlHttp = false; 
    try { 
    xmlHttp = new ActiveXObject("Msxml2.XMLHTTP") // For Old Microsoft Browsers 
    } 
    catch (e) { 
    try { 
     xmlHttp = new ActiveXObject("Microsoft.XMLHTTP") // For Microsoft IE 6.0+ 
    } 
    catch (e2) { 
     xmlHttp = false // No Browser accepts the XMLHTTP Object then false 
    } 
    } 
    if (!xmlHttp && typeof XMLHttpRequest != 'undefined') { 
    xmlHttp = new XMLHttpRequest();  //For Mozilla, Opera Browsers 
    } 
    return xmlHttp; // Mandatory Statement returning the ajax object created 
} 

var xmlhttp = new getXMLObject(); //xmlhttp holds the ajax object 

function ajaxFunction() { 
    var getdate = new Date(); //Used to prevent caching during ajax call 
    if(xmlhttp) { 

    xmlhttp.open("GET","ControlServlet?gettime=gettime" ,true); 
    xmlhttp.onreadystatechange = handleServerResponse; 
    xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded'); 
    xmlhttp.send(null); 
    } 
} 

function handleServerResponse() { 
    if (xmlhttp.readyState == 4) { 
    if(xmlhttp.status == 200) { 
     document.myForm.time.value=xmlhttp.responseText; //Update the HTML Form element 
    } 
    else { 
     alert("Error during AJAX call. Please try again"); 
    } 
    } 
} 
</script> 
</head> 
<body> 
<form name="myForm"> 
Server Time:<input type="text" name="time" /> 
<br /> 
<input type="button" onClick="ajaxFunction()" value="Click"/> 
<br /> 
</form> 
</body> 
</head> 
</html>

這裏的servlet吧：

public void doPost(HttpServletRequest request,HttpServletResponse response) throws IOException, ServletException { 
     StringBuffer requestURL = request.getRequestURL(); 
     if (request.getQueryString() != null) { 
      requestURL.append("?").append(request.getQueryString()); 
     } 
     String completeURL = requestURL.toString(); 

     if(request.getParameter("gettime")!= null && !request.getParameter("gettime").toString().equals("")){ 
     PrintWriter out = response.getWriter(); 
     Date df = new Date(); 
     out.println(df.getTime()); 
     } 
    } 

    public void doGet(HttpServletRequest request,HttpServletResponse response) throws IOException, ServletException { 
     doPost(request,response); 
    } 
}

這裏是運行上面的代碼後，你應該看到： enter image description here

來源

2013-06-20 16:51:32

如何使用jsp返回對AJAX的響應？

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