top button
Flag Notify
    Connect to us
      Site Registration

Site Registration

How to send a request to server using XMLHttpRequest Object?

+2 votes
342 views
How to send a request to server using XMLHttpRequest Object?
posted Jul 3, 2015 by Shivaranjini

Share this question
Facebook Share Button Twitter Share Button LinkedIn Share Button

1 Answer

0 votes

The XMLHttpRequest object is used to exchange data with a server.

Send a Request To a Server
To send a request to a server, we use the open() and send() methods of the XMLHttpRequest object:

xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

Method Description

open(method, url, async) Specifies the type of request

method: the type of request: GET or POST
url: the server (file) location
async: true (asynchronous) or false (synchronous)
send() Sends the request to the server (used for GET)
send(string) Sends the request to the server (used for POST)

GET or POST?

GET is simpler and faster than POST, and can be used in most cases.

However, always use POST requests when:

A cached file is not an option (update a file or database on the server).
Sending a large amount of data to the server (POST has no size limitations).
Sending user input (which can contain unknown characters), POST is more robust and secure than GET.

GET Requests

A simple GET request:

Example

xhttp.open("GET", "demo_get.asp", true);
xhttp.send();

In the example above, you may get a cached result. To avoid this, add a unique ID to the URL:

Example

xhttp.open("GET", "demo_get.asp?t=" + Math.random(), true);
xhttp.send();

If you want to send information with the GET method, add the information to the URL:

Example

xhttp.open("GET", "demo_get2.asp?fname=Henry&lname=Ford", true);
xhttp.send();

POST Requests

A simple POST request:

Example

xhttp.open("POST", "demo_post.asp", true);
xhttp.send();

To POST data like an HTML form, add an HTTP header with setRequestHeader(). Specify the data you want to send in the send() method:

Example

xhttp.open("POST", "ajax_test.asp", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send("fname=Henry&lname=Ford");

Method Description

setRequestHeader(header, value) Adds HTTP headers to the request

header: specifies the header name
value: specifies the header value
The url - A File On a Server
The url parameter of the open() method, is an address to a file on a server:

xhttp.open("GET", "ajax_test.asp", true);

The file can be any kind of file, like .txt and .xml, or server scripting files like .asp and .php (which can perform actions on the server before sending the response back).

Asynchronous - True or False?

AJAX stands for Asynchronous JavaScript and XML, and for the XMLHttpRequest object to behave as AJAX, the async parameter of the open() method has to be set to true:

xhttp.open("GET", "ajax_test.asp", true);

Sending asynchronous requests is a huge improvement for web developers. Many of the tasks performed on the server are very time consuming. Before AJAX, this operation could cause the application to hang or stop.

With AJAX, the JavaScript does not have to wait for the server response, but can instead:

execute other scripts while waiting for server response
deal with the response when the response ready
Async=true
When using async=true, specify a function to execute when the response is ready in the onreadystatechange event:

Example

xhttp.onreadystatechange = function() {
  if (this.readyState == 4 && this.status == 200) {
    document.getElementById("demo").innerHTML = this.responseText;
  }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

You will learn more about onreadystatechange in a later chapter.
Async=false
To use async=false, change the third parameter in the open() method to false:

xhttp.open("GET", "ajax_info.txt", false);

Using async=false is not recommended, but for a few small requests this can be ok.

Remember that the JavaScript will NOT continue to execute, until the server response is ready. If the server is busy or slow, the application will hang or stop.

Note: When you use async=false, do NOT write an onreadystatechange function - just put the code after the send() statement:

Example

xhttp.open("GET", "ajax_info.txt", false);
xhttp.send();
document.getElementById("demo").innerHTML = xhttp.responseText;
answer Sep 26, 2016 by Manikandan J
Similar Questions
+2 votes

I am working on an app with Java and Struts2.
I fetch all data from my DB to an Object(eg.XXX) List at once while Tomcat(7) is coming up. Then according to user selection, I am trying to get data from the same List by doing Ajax call. But it is failing and getting an empty list. I don't want to use session.
Can I use and global object list for this?? An Example will be more helpful.

+3 votes

The browser i am using is google chrome. I also tested it in internet explorer and firefox too. But result is same. The post method is working fine for the same rest endpoint, but when i am trying to execute delete method on the same endpoint, it's throwing error with http status code 209 with the reason "failed to form full key for endpoint".

This is how the i am sending the delete request.

$.ajax({
             url: url,
             type: "DELETE",
             dataType:"json",
             data: JSON.stringify(data),
             crossDomain: true,
             contentType:"application/json",
             success:function(result){console.log("Success " + console.log(JSON.stringify(result)))},
             error:function(error){
                        alert("Error Deleting Service.See the console log.");
                        console.log(JSON.stringify(error));
                        },
             }).then(function(response){
                   console.log(JSON.stringify(response));
             });

The POST method:

 $.ajax({
                 url: url,
                 type: "POST",
                 dataType:"json",
                 data: JSON.stringify(data),
                 crossDomain: true,
                 contentType:"application/json; charset=utf-8",
                 success:function(){console.log("Success")},
                 error:function(error){console.log(JSON.stringify(error))},
                 }).then(function(response){
                      console.log(JSON.stringify(response));
                 });

Need help. I have already spent the day on this but not able to find the solution for this. I also tried to append the data params with the url but that too didn't worked.

...