I want to make an ajax call inside my jsp file which calls processAction method of a portlet, based on the success message from processAction method i need to make another call to serveResource method of portlet,please provide some examples..
In portlets, processAction() methods are automatically followed by render method and hence ajax response will get embedded with HTML fragment generated by render method. So writing ajax in portlets is a bit tricky.
Have a look at this blog of mine.
http://ajax-and-portlets.blogspot.com/2011/09/ajax-best-practice-in-portlets.html
It gives an insight view of what's the best practice to implement ajax in portlets (for both JSR-168 and JSR-286 portlets).
In case you want sample portlets, you can contact me through the contact details from the blog. I'll be happy to help you.
Thanks
Jignesh
This question worked for me.
Basically, the Controller
#Controller
#RequestMapping("VIEW") // VIEW mapping (as opposed to EDIT)
public class MyPortlet {
#RenderMapping
public String handleRenderRequest(RenderRequest request, RenderResponse response) {
return "defaultRender";
}
#ResourceMapping("myURL")
public void handleMyResource(ResourceRequest request, ResourceResponse response) {
OutputStream outStream;
try {
outStream = response.getPortletOutputStream();
ObjectMapper mapper = new ObjectMapper();
mapper.writeValue(outStream, "Hello world!");
} catch (IOException ex) {
// TODO : Do something with errors.
}
}
}
And the JSP:
<portlet:resourceURL id="myURL" var="myURL"/>
<script type="text/javascript">
var urlink = "<%= myURL %>";
$.ajax({
url: urlink,
cache: false,
type: "POST",
success: function(jsondata) {
console.log(jsondata);
}
});
</script>
based on the success message from processAction method
That's not the right way to do it.
On calling portlet action URL in response you get usual render response, so you'll get page with all the portlets.
Instead you should use Portlet 2.0 resource serving feature, and return your response as a resource.
You can check out my portlet which has examples for both serveResource and processAction methods calling.
Ajax Jquery Portlet
Related
I am calling a spring controller method from dojo xhrPost and I need to redirect to a new html from the controller method. But controller returns back to the dojo in javascript instead of moving to a new page.
my javascript:
var xhrArgs={
url:"/tradeIn/submitTradeIn",
postData:dojo.toJson(tradeInDetails),
handleAs:"text",
headers:{"Content-Type":"application/json"}
/*load:function(data){
console.log(data);
label=data;
if(data =="fail"){
location.reload(true);
window.location="/Fail";
}
else{
window.location="/success";
}
}*/
}
var deferred=dojo.xhrPost(xhrArgs);
my spring controller:
#RequestMapping(value="/tradeIn/submitTradeIn", method = {RequestMethod.POST})
public String submitTradeIn(#RequestBody TradeInDetails tradeDetails) throws UnsupportedEncodingException{
List<byte[]> labelList=new ArrayList<byte[]>();
ShippingInfo shippingInfo=new ShippingInfo();
shippingInfo.setAddress1(tradeDetails.getCustomerDetails().get(0).getAddress1());
shippingInfo.setAddress2(tradeDetails.getCustomerDetails().get(0).getAddress1());
shippingInfo.setCity(tradeDetails.getCustomerDetails().get(0).getCity());
shippingInfo.setCompany(tradeDetails.getCustomerDetails().get(0).getCompany());
shippingInfo.setDayPhone(tradeDetails.getCustomerDetails().get(0).getDayPhone());
shippingInfo.setEmail(tradeDetails.getCustomerDetails().get(0).getEmail());
shippingInfo.setEvePhone(tradeDetails.getCustomerDetails().get(0).getEvePhone());
shippingInfo.setFirstName(tradeDetails.getCustomerDetails().get(0).getFirstName());
shippingInfo.setLastName(tradeDetails.getCustomerDetails().get(0).getLastName());
shippingInfo.setState(tradeDetails.getCustomerDetails().get(0).getState());
shippingInfo.setZip(tradeDetails.getCustomerDetails().get(0).getZip());
shippingInfo.setCountry(tradeDetails.getCustomerDetails().get(0).getCountry());
List<ReturnRequestLabel> label;
List<TradeInClubs> tradeInList1= new ArrayList<TradeInClubs>();
for(ClubDetails cl: tradeDetails.getClubDetails()){
TradeInClubs tradeInclubs1=new TradeInClubs();
tradeInclubs1.setClubMaterial(cl.getShaftType());
tradeInclubs1.setClubType(cl.getClubType());
tradeInclubs1.setManufacturer(cl.getClubManufacturer());
tradeInclubs1.setModel(cl.getClubModel());
tradeInclubs1.setTradeInValue(cl.getTradeInPrice());
tradeInList1.add(tradeInclubs1);
}
try{
ReturnFedexLabel returnFedexLabel = new ReturnFedexLabel();
label=returnFedexLabel.fetchFedexLabel(tradeInList1, shippingInfo);
byte[] labelImageData;
String fedexLabelNumber=null;
for(ReturnRequestLabel rl: label){
labelImageData=rl.fedexReturnLabel.imageData;
labelList.add(labelImageData);
fedexLabelNumber=rl.trackingNumber;
}
File f=new File("label.jpg");
BufferedImage img=ImageIO.read(new ByteArrayInputStream(labelList.get(0)));
ImageIO.write(img,"JPEG",f);
int id=tradeInDao.insertQuery(shippingInfo,tradeInList1,fedexLabelNumber);
byte[] pdfData=fedexLabelToPdf.printFedexLabel(labelList);
emailTradeIn.emailTradeInDetails(pdfData,tradeDetails.getCustomerDetails().get(0).getEmail(),tradeInList1,id);
System.out.println("here");
} catch (Exception e){
logger.error(e.getMessage());
return "fail";
}
return "success";//Base64.encodeBase64String(labelList.get(0));
}
it is not moving to success page. it just stays in the current page
Check and see what deferred is, it is probably "success".
From http://dojotoolkit.org/reference-guide/1.7/quickstart/ajax.html:
dojo.xhrPost xhrPost will create an Ajax request using the HTTP POST
method and is usually used to submit data to a service. It returns
data to a callback. The callback is defined as a member of the object
used to create the request (the property-bag), or by using the
dojo.Deferred.then() method.
For complete details and examples, see the dojo.xhrPost documentation.
if you need to change the page from the controller it shouldn't be an ajax request. If you can change it from the client side then you can continue with the ajax request and respond accordingly.
Scenario
Lotus Domino form with a button that made an Ajax call to an Xpage, that do some stuff (read a properties file).
Framework: prototype.js
Button code:
var now = new Date()
var n = $H({
........
now: now.getTime()
});
var url = "/" + $F("path") + "/myxpages.xsp";
var myAjax = new Ajax.Request(
url,
{
method: 'post',
parameters: n.toQueryString(),
onComplete: function(response) {
ajaxResult = response.responseText;
}
});
Xpage
myxpages.xsp has this SSJS code on afterPageLoad event
var request = facesContext.getExternalContext().getRequest();
var response = facesContext.getExternalContext().getResponse();
response.setHeader("Expires", -1);
response.setHeader("Cache-Control", "no-cache");
com.org.MyGetProperties.readProperties(request,response);
MyGetProperties class
This class is deployed in WebContent/WEB-INF/classes
public class MyGetProperties {
static PrintWriter out = null;
public static synchronized void readProperties(HttpServletRequest request, HttpServletResponse response) throws Exception {
try {
*(DO SOME STUFF HERE)*
out = new PrintWriter(response.getWriter());
// return result
out.println("OK");
} catch (Exception e) {..}
}
}
Sometimes ajaxResult variable in ajax response call is empty, sometimes is "OK", as expected (seems to be something related to cache, but i think i've managed it correctly).
The behavior is different on different production server, i don't know if depends on server configurations.
Could be a PrintWriter problem?
Short answer: don't. Long answer: use the Ajax control. You put that on your page and your URL changes to myxpages.xsp/nameyougavetheajaxcontrolproperty
This way you can be sure not to run foul of any cached result or pending operation. It also has a property where you can specify a Java class directly. That class extends (need to Google that - answered it on SO before) which gives you direct access to request/response
Update:
You need to tell that you are done:
facesContext.responseComplete();
See my original post on XAgents, the revision and some thought on testing.
I'm trying to extract form data from within the OnActionExecuting method of an ASP.NET Web API Action Filter so I can write it to a log file.
My code looks like this:
public override void OnActionExecuting(System.Web.Http.Controllers.HttpActionContext ctx)
{
if (ctx.Request.RequestUri.Query.StartsWith("admin/")) return;
try
{
_loggingService.LogRequest(new RequestLogModel
{
Handler = ctx.ControllerContext.ControllerDescriptor.ControllerName,
Uri = ctx.Request.RequestUri.AbsoluteUri,
RequestType = ctx.Request.Method.Method,
RequestFrom = ctx.Request.RequestUri.Host,
QueryString = ctx.Request.RequestUri.Query,
FormData = // Need to get form-data here
});
}
catch { }
I can't seem to work out how to get the form data from the request. Can anyone help?
You should be able to get the form data in the request's content.
For more information on reading the multipart/form-data content using ReadAsMultipartAsync(...), check out this tutorial, Sending HTML Form Data.
Investigating the Web API as part of an MVC 4 project as an alternative way to provide an AJAX-based API. I've extended AuthorizeAttribute for the MVC controllers such that, if an AJAX request is detected, a JSON-formatted error is returned. The Web API returns errors as HTML. Here's the AuthorizeAttribute that I'm using with the MVC controllers:
public class AuthorizeAttribute: System.Web.Mvc.AuthorizeAttribute
{
protected override void HandleUnauthorizedRequest(AuthorizationContext filterContext)
{
base.HandleUnauthorizedRequest(filterContext);
filterContext.Result = new RedirectToRouteResult(
new RouteValueDictionary
{
{ "area", "" },
{ "controller", "Error" },
{ "action", ( filterContext.HttpContext.Request.IsAjaxRequest() ? "JsonHttp" : "Http" ) },
{ "id", "401" },
});
}
}
How could I reproduce this to provide equivalent functionality for the Web API?
I realize that I need to extend System.Web.Http.AuthorizeAttribute instead of System.Web.Mvc.AuthorizeAttribute but this uses an HttpActionContext rather than an AuthorizationContext and so I'm stuck by my limited knowledge of the Web API and the seemingly incomplete documentation on MSDN.
Am I even correct in thinking that this would be the correct approach?
Would appreciate any guidance.
To get the equivalent functionality in a Web API filter you can set the HttpActionContext.Response property to an instance of HttpResponseMessage that has the right redirect status code and location header:
protected override void HandleUnauthorizedRequest(HttpActionContext actionContext) {
var response = new HttpResponseMessage(HttpStatusCode.Redirect);
response.Headers.Location = new Uri("my new location");
actionContext.Response = response;
}
I would very much go with Marcin's answer - at the end of the day, he has written the code!
All I would add is that as Marcin is saying, your best bet is to have a dedicated controller to return the errors as appropriate - rather than setting the response code 401 with JSON content in the attribute.
The main reason is that Web API does the content-negotiation for you and if you want to do it yourself (see if you need to serve JSON or HTML) you lose all that functionality.
I have a login form (login.jsp) with two input fields, username and password.
I am using POST via Ajax to access the login servlet.
I want the user to login, and if the login is successful, be redirected to another page called 'search.jsp'. If unsuccessful, a 'login failed' message is returned as the Ajax responseText to be inserted into a paragraph in the 'login.jsp' page.
I have everything working, my login servlet accesses the database via a separate bean, and an object of that bean is returned with its properties ready to use. So all is good to there.
But, after the username and password pass muster with the database, I'm then using RequestDispatcher to forward to the new landing page (search.jsp).
Here is my doPost()
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String username,password;
username = request.getParameter("p");
password = request.getParameter("q");
try {
LoginService ls = new LoginService(username,password);
User user = ls.getUserDetails();
if(user.getUsername()!=null && user.getPassword()!=null){
FormService filler = new FormService();
Form fields = filler.getFields();
request.setAttribute("user",user);
request.setAttribute("fields1",fields);
request.setAttribute("fields2",fields);
request.setAttribute("fields3",fields);
HttpSession session = request.getSession(true);
//set attribute for the session
session.setAttribute("user",user.getUsername());
//Now, the RequestDispatcher.forward() is not forwarding to the new page!
//The whole 'search.jsp' page is being stuffed back into the 'login.jsp' page
RequestDispatcher rd = request.getRequestDispatcher("search.jsp");
rd.forward(request,response);
return;
}
else{
PrintWriter out = response.getWriter();
out.println("login failed!");
return;
}
} catch (Exception e) {
e.printStackTrace();
}
}
But instead of forwarding the request and response to the new jsp page 'search.jsp', the whole search.jsp page is being stuffed back into the the original login.jsp page - in the html element which holds the Ajax responseText in when login fails.
The forward() method in the servlet works when the servlet is called from the form action attribute, but not when the servlet is called the javascript file containing the Ajax code.
But instead of forwarding the request and response to the new jsp page 'search.jsp', the whole search.jsp page is being stuffed back into the the original login.jsp page - in the html element which holds the Ajax responseText in when login fails.
That's indeed the expected behaviour. You're handling the request/response using JavaScript. Your JavaScript code has retrieved the response of search.jsp as responseText and is putting it in the HTML element.
You need to change this approach. You need to let the response return the necessary data which sufficiently informs JavaScript so that it can handle the response properly. A commonly used data format for this is JSON.
Something like
response.setContentType("application/json");
if (user != null) {
// ...
response.getWriter().write("{ 'success': true, 'location': 'search.jsp' }");
} else {
response.getWriter().write("{ 'success': false, 'message': 'Unknown login' }");
}
and in JS:
var responseJson = eval('(' + xhr.responseText + ')');
if (responseJson.success) {
window.location = responseJson.location;
} else {
document.getElementById('message').innerHTML = responseJson.message;
}
If you want to handle this unobtrusively, so that the same servlet is reuseable on normal (non-ajax) HTTP requests (so that your webapp still works when the client has JS disabled!) then you could check if the X-Requested-With header equals to XmlHTTPRequest.
if ("XMLHttpRequest".equals(request.getHeader("X-Requested-With")) {
// Handle ajax response (e.g. return JSON data object).
} else {
// Handle normal response (e.g. forward and/or set message as attribute).
}
See also:
How to use Servlets and Ajax?
Simple calculator with JSP/Servlet and Ajax