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return json object using servlet
Change return "main"; to return null; After that you will be able to get ajax response and alert() on the same page; Instead of alert you can redirect user to another page now if needed by doing document.location="anotherpage";

Categories : Jquery

get value separately from json object in java servlet
In your servlet, do the following in the do[Get,Post]() method JSONObject jsonObject = JSONObject.fromObject( request.getQueryString() ); String username= jsonObject.get( "username" ); String password= jsonObject.get( "password" ); String type= jsonObject.get( "type" ); You also need the remove the following line from your JS. request = JSON.parse(data);

Categories : Java

Tomcat servlet acting strange with json object
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("application/json"); PrintWriter out = response.getWriter(); Gson gson = new Gson(); LocationTypes locTypes = new LocationTypes(); String json = gson.toJson(locTypes); response.setContentType("application/json"); out.print(json); out.flush(); } try setting content-type as above

Categories : Json

send and receive byte array form servlet to another servlet via ur
URLConnection conn = new URL(URL to send input stream").openConnection(); conn.setDoOutput(true); OutputStream outs = conn.getOutputStream(); outs.write(pdfBytes); outs.flush(); outs.close(); on other side get that input stream from request request.getInputStream();

Categories : Java

JSP - Servlet "How to send table data from jsp to servlet"?
You need to create a hidden input type and assign the value of td tag to it and access its value in servlet using request.getparameter("hiddenInputName") e.g. <td>someData</td><input type = "hidden" name = "someName" value = "someData"> Then after submitting form using post method you can access the value in servlet as request.getParameter("someName");

Categories : Jsp

How to send json object to C# web server
data option for .ajax expects a name value pair string, or an object data: { "myjson": JSON.stringify({ person:{ firstName: "Denny" }}) }, //OR data: "myjson="+JSON.stringify({ person:{ firstName: "Denny" }}), //Or just send the data values and retrieve in the way you get GET or POST variables in C# data: { person:{ firstName: "Denny" }},

Categories : C#

How to send JSON object to rest webservice?
I cannot see on your post where are you doing the call to the database. All I can see is that you are creating a Student and then, from there, creating a view (assuming this method is being called). The error you are getting is related to the fact that, at some point, you are calling the database to enter a new row in (probably) your student Model. If you have got hibernate, can you make sure that you are not calling .save or .update at any point between the call you get for the student and before you call setName?

Categories : Java

How to send json object to Rabbitmq server
You can try to do it using Jackson. See this tutorial for example. Other alternatives: google-gson json-lib flexjson

Categories : Json

send complex object json to webservice with jquery
Ok I found the problem public class PaymentForm { int id { get; set; } string name { get; set; } } I forgot PUBLIC :-( public class PaymentForm { public int id { get; set; } public string name { get; set; } }

Categories : Jquery

How to send data as json object in post method of Rho::AsyncHttp
First, when you set the :body => #{data}, you are converting the json data to string doing #{} Second, you need to set the :header to { "Content-Type" => "application/json; charset=utf-8", "Accept" => "application/json" }. This makes the post method accept the data as json object. So you need the below code, data = { :query1 => "this demo" , :query2 => "another demo" } # Any data to be passed result = Rho::AsyncHttp.post( :url => "https://www.myserverapi.com/myapi", :header => { "Content-Type" => "application/json; charset=utf-8", "Accept" => "application/json" }, :body => Rho::JSON.parse(data) ) Hope this helps you.

Categories : Json

send json array object as a parameter to a php service from android
This is how I believe it should be done: URL url = new URL("//ip of your PHP function on your web server"); HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection(); StringBuilder postDataBuilder = new StringBuilder(); postDataBuilder.append(URLEncoder.encode("005", "UTF-8")); urlConnection.setDoOutput(true); urlConnection.setRequestMethod("POST"); urlConnection.setUseCaches(false); urlConnection.setConnectTimeout(10000); urlConnection.setReadTimeout(10000); urlConnection.setRequestProperty("Content-Type","application/json"); urlConnection.connect(); JSONObject jsonParam = new JSONObject(); jsonParam.put("item 1", "value 1"); //Add other pairs here, as you are doing OutputStreamWriter out = new OutputStreamWriter(urlConnection.getOutputStream()); out.write(jsonParam.toSt

Categories : PHP

How do I send the results of a javascipt method (JSON object) to the code-behind on asp.net webform?
I see two immediate options, depending upon what you want to do in the server-side code: Manually invoke __doPostBack JavaScript function to cause a server-side postback. I would recommend this approach if you want to interact with the ASP.NET server controls on the page. Here is another StackOverflow question that details how to send Multiple parameters in __doPostBack. Call an ASP.NET AJAX Page Method, which is essentially a stand-alone web service hosted inside of an ASP.NET page. I would recommend this approach if you just want to invoke some server-side logic (i.e. save or retrieve data from the database), but do not not need an instance of the page itself, as this is an asynchronous operation and none of the page controls will be available. This is great for just sending data t

Categories : C#

How to send JSON object/string (as 'POST') to a webservice using $resource in angularjs
The good news is that this should be fairly straight forward. You have the right idea, you just need to extend the resource object as you have done with "post" method, and when you call it - pass in your JSON object. Angular will then append the passed in JSON object as post parameters automatically. It's a good idea (I've read), to create this User resource as a service/factory so that you can inject it into your controllers to abstract the calls to the server. As an example - something I have done would be to create the service like so (with a dependency on the angular $resource): var myApp = angualar.module('myApp', []); myApp.factory('UserService', ["$resource", function UserService($resource) { var UserService = $resource('/rest/user', {}, {

Categories : Json

Send data from servlet to jsp
1) You need to add import statements at top of the JSP. Example: <%@ page import="java.util.List" %> 2) It is NOT good practice to have Java code directly embedded in JSP Read more here on SO Wiki

Categories : Java

the attribute send by jsp is nowhere in servlet
This is the way you would retrieve parameters in servlet from following URL, www.someServer.com/somePage.jsp?monitorScreen=true Then in doGet method of servlet, String value = request.getParameter("monitorScreen");

Categories : Jsp

how to send ArrayList from jsp to servlet
How to send arrayList from JSP to Servlet. Is it possiple Yes definitely it is possible, you need to stuff the arrayList object in the request object. Retrieve the list in servlet. As you have not provided information whether you are using scriptlets or jstl, I'm assuming the scriptlets. In your JSP, request.setAttribute("arrayList", yourListObject); RequestDispatcher rd = request.getRequestDispatcher("servletUrlPattern"); rd.forward(request, response); In your servlet, you can access the list using List yourList = (List)reqeust.getAttribute("arrayList"); To call the servlet from your JSP file, <form action = "yourServletUrl" method = "POST"> //everything in the form here. <input type = "submit" name = "submit"> </form> After pressing s

Categories : Jsp

How can I send http request to another servlet
You could send a Get request using Java; URL url = new URL("http://localhost:8080/Mapping/App2"); HttpURLConnection conn = (HttpURLConnection)url.openConnection(); conn.setRequestMethod("GET"); conn.connect(); Alternatively you should probably configure App2 so that it's action is handled by a separate class or a method accessible to both servlets.

Categories : Java

Servlet.service() for servlet [Faces Servlet] in context with path threw exception [null] with root cause
The error is well detailed, above the exception you mention com.sun.faces.application.view.FaceletViewHandlingStrategy handleRenderException SEVERE: Error Rendering View[/cECI/domainReport.xhtml] java.lang.NullPointerException You should give a look into that xhtml so it can be rendered correctly Also, you can create a custo lo4j.properties at /src folder, to obtain extra information from ICEFaces core (in the example is set to WARN, try with INFO, or DEBUG, until you find more information on your problem) log4j.rootLogger=INFO log4j.appender.stdout=org.apache.log4j.ConsoleAppender log4j.appender.stdout.layout=org.apache.log4j.PatternLayout log4j.appender.stdout.layout.ConversionPattern=[%d{dd MMM yyyy HH:mm:ss}] %-5p [%F:%L -- %M] - %m%n log4j.logger.com.icesoft.faces.application.D

Categories : Java

How to send PDF file data as a response using Servlet?
package com.javatpoint; import java.io.*; import java.util.*; import javax.servlet.*; import javax.servlet.http.*; import com.darwinsys.spdf.PDF; import com.darwinsys.spdf.Page; import com.darwinsys.spdf.Text; import com.darwinsys.spdf.MoveTo; public class ServletPDF extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException { PrintWriter out = response.getWriter(); response.setContentType("application/pdf"); response.setHeader("Content-disposition","inline; filename='javatpoint.pdf'"); PDF p = new PDF(out); Page p1 = new Page(p); p1.add(new MoveTo(p, 200, 700)); p1.add(new Text(p, "www.javatpoint.com")); p1.add(new Text(p, "by Sonoo Jaiswal")); p.add(p1); p.setAuthor("Ian F. Darwin"); p.writePDF(); } }

Categories : Java

Send parameter from script to servlet with POST
It sounds like you want to post that form. If so: function redireccionar() { document.form1.method = "POST"; // Don't need this if it's in the markup document.form1.action = "http://lnxntf05:8080/MyMaver/ServletTipoPapel"; // Could put this in the markup as well document.form1.submit(); } setTimeout(redireccionar, 20); But you'll need to not disable the form elements (remove document.form1.CLAVEUSU.disabled=true and such) because disabled form elements aren't sent with forms. Note that it's usually best not to use strings with setTimeout. I've used a function reference above.

Categories : Javascript

Not able to send request to Servlet doPost method from HttpClient
Your contentType is wrong, to upload a file to a web server you need to specify the multipart format. See http://stackoverflow.com/a/1068132/305116 for a problem like yours, and http://evgeny-goldin.com/blog/uploading-files-multipart-post-apache/ for a little tutorial. So in your main function you need something like this for it to work: public static void main(String[] args) { HttpPost post = new HttpPost("http://localhost:8080/ServletExample/SampleServlet"); MultipartEntity entity = new MultipartEntity(); entity.addPart( "someXMLfile", new StringBody(generateNewXML(), "application/xml", Charset.forName( "UTF-8" ))); post.setEntity(entity); HttpClient client = new DefaultHttpClient(); HttpResponse response = client.execute(post); }

Categories : Java

How can I send the selected information in an html page(JSP) to a servlet?
Have you tried this? <form action="ExportDB" target="_blank"> <select name="day"> <option value=""></option> <option value="01">01</option> <option value="02">02</option> ... <option value="30">30</option> <option value="31">31</option> </select> <select name="month"> <option value=""></option> <option value="01">01</option> <option value="02">02</option> ... <option value="11">11</option> <option value="12">12</option> </select> <select name="year"> <option value=""></option> <option value="2013">2013</option> ... <option value="2029">2029<

Categories : Java

eclipse servlet java.lang.ClassNotFoundException: javax.servlet.Servlet
Is that error happing during the compilation or during the run? Did you configure Eclipse to point to the Tomcat server? In order to run a web server in eclipse using tomcat, you need to tell eclipse about tomcat first. Here is where you ca find the instruction. http://www.coreservlets.com/Apache-Tomcat-Tutorial/tomcat-7-with-eclipse.html

Categories : Java

eroor while deleting entry and send resultset from servlet to jsp page
Your code is a terrible mess and you create several memory leaks here. Use a DataSource factory like Tomcat JDBC Pool and Apache Commmons DbUtils to focus on your problem. Moreover, you do not perform a conn.commit() after the delete. You are in autocommit mode. You should never work in autocommit mode. Your pass a result set which has already been closed by the connection.

Categories : Jsp

Servlet that receives a XML based request and then make a new XML file to send back as the response
You probably want to do everything in the doPost() method. Just one of doGet or doPost will be called, for a given HTTP request, depending on if the caller specified GET or POST in their request. Your creation of the XML response looks basically ok. That is the general approach anyway, write the result XML to the response writer. If this is for production code, and not just a learning exercise, then you should use a library to create the XML, not just manually build it from strings. See "HOWTO Avoid Being Called a Bozo When Producing XML" http://hsivonen.iki.fi/producing-xml/ As far as parsing the incoming request: BufferedReader reader = request.getReader() Use that to read the characters of the incoming XML, and pass them to your XML parser.

Categories : Java

how to make json object from an element in json object with unique values in underscore js
var col = [ { "MerchantName": "Fashion and You", "BrandList": " Nike, Fila", "MerchantImage": "Fashion-You-medium.jpeg" }, { "MerchantName": "Fashion and You", "BrandList": " Levis, Fasttrack, Fila", "MerchantImage": "Fashion-You-medium.jpeg" }, { "MerchantName": "ebay", "BrandList": "Nokia,HTC,Samsung", "MerchantImage": "ebay.jpeg" }, { "MerchantName": "amazon", "BrandList": "Apple,Dell,Samsung", "MerchantImage": "amazon.jpeg" }, { "MerchantName": "amazon", "BrandList": " pepe jeans, peter england, red tape", "MerchantImage, Fila": "amazon.jpeg" } ]; var brands = []; //get unique brands _.each(col, function(i){ brands = _.union

Categories : Javascript

how to check form hidden field is empty or not and then only send respected request parameter to servlet
wrap your input in a form like this: <form name="section2-next" id="section2-next"> <input type="text" id="data-1-Id"/> <input type="text" id="data-2-Id"/> <input type="text" id="data-3-Id"/>><input type="text" id="data-4-Id"/> <input type="submit"value="submit!"/> <!-- dont forget the input of type submit--> </form> then use $("#section2-next").submit(function(event){ event.preventDefault();// dont forget this line.. //instead of $("#section2-next").click(function(){ next // for your dataString var dataString = {param1 : $("#data-1-Id").val() || '', param2 : $("#data-2-Id").val() || ''};// up until param6 and // for var parsed = JSON.parse(data); var parsed = $.parseJSON(data); finally, // in

Categories : Javascript

Create a empty JSON object from an existing JSON object array
You can use an object literal to store whatever you want. It is just a bag of properties (i.e. name) and values. e.g. var order = {}; Then an array literal could be used to hold the orders. e.g var orders = []; orders.push(order); But it would be just as easy to use another object literal with the id as a property. But it seems like you want some sort of validation. Perhaps something to manage the order data and handle the validation, etc. Like so: orderManager.dataStore = { _data: {}, //_redundantData = []; //could easily store in an array if id isn't unique get: function (id) { return this._data[id]; }, getAll: function () { return this._data; }, set: function (id, order) { validateOrder(order); this._data[id] = order; }, clear: function (id)

Categories : Javascript

How to send parameter from a form to Servlet (date of Birth) (select day, month, year, option) Java
Once you have your values in your servlet you can use SimpleDateFormat#format() to format a Date into a String in a certain pattern. String newstring = new SimpleDateFormat("yyyy-MM-dd").format(date); System.out.println(newstring); // 2013-01-15 Also, your html code must be inside a form, something like this: <form:form method="post" action="addContact.html"> <!-- html fields.. --> </form:form> In your servlet, try overriding doGet and doPost methods. Take a look at: http://stackoverflow.com/tags/servlets/info

Categories : Java

get json array from servlet using ajax
See, if you want to pass that from servlet to jsp you need not to make a request (ajax), since servlet and jsp both exists at server side.You just set that json array as a request attribute or session attribute and get it in jsp and display(with loop).NO need of ajax there. If You need to fetch the data with synchronous call: In servlet PrintWriter out = response.getWriter(); out.println(htags); I won't clutter SO with another full example,see this SO post:How to send JSON array from server to client, i.e. (java to AJAX/Javascript)?

Categories : Java

Reading JSON values in servlet
You do not need to stringify the data ... just send it as a plain old javascript object ... by specifying datatype as json ... jquery will figure out how to pack the data in the request No need to change anything on server side So if your AJAX call becomes: $.ajax({ url: 'dvsds', type: 'POST', dataType: 'json', data: jsondata, error:function(){ alert("error occured!!!"); }, success:function(data){ alert(data.message); } }); Retrieve them in the servlet as String fname = request.getParameter("fullname"); String mobileno = request.getParameter("mobileno"); I think you will need to be careful about case sensitivity EDITED I see that you Can you change your script to be as follows. <script type="text/javascript

Categories : Ajax

ClassNotFoundException in JSON Parsing by servlet
The most likely explanation is that the JAR that defines the Apache NestableRuntimeException is missing from your webapp's WAR file. According to http://www.findjar.com, there are lots of JAR file that could supply that dependency, but the most likely one is one of the commons-lang-x.x.jar files; e.g. commons-lang-2.3.jar If you are building your WAR file using Maven, just add the relevant dependency to the POM file. Otherwise, download the JAR by hand and add it to the "/WEB-INF/lib" directory of your WAR file.

Categories : Java

how to send media file to wcf rest json service and get string return response in json format
To send a media file (or any arbitrary file, for that matter) to WCF, you need to have an operation where the only parameter in the request body is of type Stream. That means that you can have other parameters to the operation, but they need to be passed via the URI (using the UriTemplate property of the [WebInvoke] attribute) - see more information in the post at http://blogs.msdn.com/b/carlosfigueira/archive/2008/04/17/wcf-raw-programming-model-receiving-arbitrary-data.aspx. In your example, you'd have something like the code below: [OperationContract] [WebInvoke(Method = "POST", UriTemplate = "RegisterUser?emailId={EmailID}&name={Name}&mobile={Mobile}&IMEI={IMEI}", BodyStyle = WebMessageBodyStyle.WrappedRequest, RequestFormat= WebMessageFormat.Json, ResponseFor

Categories : Android

Receving send json to server and recive json response
It may be already answered in How to decode JSON values in my Android Aplication? BUT, if you want to parse JSON Data from the App to the Webserver you may want to use the flag true. $post = json_decode(filter_input(INPUT_POST, 'data'), true); foreach ($post as $obj) { echo $obj["yourobj"]; } If you have problems parsing the data from the webserver to the app, force it to be an object by using json_encode($data, JSON_FORCE_OBJECT); if you want to parse an array use json_encode(array("data", $yourdata)); Please explain what you want and the error you get for a proper solution matching your case.

Categories : Android

Getting json response from servlet and display in jsp table?
I could not see "dataType" in Your Ajax request like: dataType: 'json' try adding it. so, it will become: $.ajax({ type: "GET", url: "./dataFetchController", success: function(responseJson) { //alert and success handler }, dataType: 'json', error : function(){ // error handler }); dataType is the type of data that you're expecting back from the server

Categories : Java

Phantomjs :: POST Data as json to a Servlet not working
Servlets deal with simple request, so they only know how to parse (natively) HTTP parameters, either GET parameters from the URL, or POST parameters sent as application/x-www-form-urlencoded. Newer versions of the Servlet specification can also read multipart/form-data. But there's nothing about JSON mentioned either in the Servlet or the HTTP specifications. So you must use a library that knows how to parse JSON and make the resulting object available in the Velocity context.

Categories : Javascript

Order JSON object alphabetical and return to JSON object
write a compare function function compare(a,b) { if (a.name < b.name) return -1; if (a.name > b.name) return 1; return 0; } then sort your data var cities = {}; $.getJSON('http://mypage.com/json/cities.php', function(data){ data.sort(compare); $.each(data, function (k, vali) { cities[vali.cid] = vali.name; }); });

Categories : Jquery

How to access same ServletContext object in another servlet?
my question is, for accessing this object i have to create ServletContext object and by using context.getAttribute(arg1,arg2) ,will i get value. or there is another value to do this. You will get value by creating context object same as above in the another servlet. According to java docs There is one context per "web application" per Java Virtual Machine.

Categories : Java

Form submission through ajax and reply from servlet through JSON without page refresh
You need to specify dataType of your response. $.ajax({ dataType: 'json', ... As per documentation, if none is specified, jQuery will try to infer it based on the MIME type of the response. As I see from your servlet code, you've specified wrong MIME type for your response : " text/html". So you didn't leave a chance to jQuery to "guess" that your response is in json format. Also, it's good way to figure out what's going on using console.log in your success method: success: function(msg){ console.log(msg); Also see getJSON function I am not able to track if the success function is called or not....can u plz let me know how to do it....i am not able to use console.log As for this comment, you can always use browser console to track your request status, look at request an

Categories : Javascript

JSON.NET: Serialize json string property into json object
You need a converter to do that, here is an example: public class RawJsonConverter : JsonConverter { public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer) { writer.WriteRawValue(value.ToString()); } public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer) { throw new NotImplementedException(); } public override bool CanConvert(Type objectType) { return typeof(string).IsAssignableFrom(objectType); } public override bool CanRead { get { return false; } } } Then decorate your class with it: public class Foo { public int Id; [JsonConverter(typeof(RawJsonConverter))] public string RawData; } Then, when you use:

Categories : C#



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