XMLHttpRequest();

[ravi1021]

Hi Team ,

Can you please guide me how to call

var client=new XMLHttpRequest();

from Script Include, i have few parameters to be added to api call, so i cant use Rest Outbound here,

please suggest me

[Ankur Bawiskar]

Hi Ravi,

If this is a URL api then you can try below code in background script

var httpclient = Packages.org.apache.commons.httpclient;

var HttpClient = httpclient.HttpClient;

var UsernamePasswordCredentials = httpclient.UsernamePasswordCredentials;

var AuthScope = httpclient.auth.AuthScope;

var PostMethod = httpclient.methods.PostMethod;

// consider end point url as http://example.com/p.php

var client = new HttpClient();

var postMethod = postMethod = new PostMethod('endpoint url');

postMethod.addParameter("A", "Test");

postMethod.addParameter("B", "Test1");

postMethod.addParameter("C", "Test User8");

postMethod.addParameter("D", "TestUser8@example.com");

// these parameters will be replaced in the URL as below

http://example.com/p.php?A=Test&B=Test1&C=Test User8&D=TestUser8@example.com

var status = client.executeMethod( postMethod );

var result = (status + "\n" + postMethod.getResponseBodyAsString()).replace(/</g, "&"+"lt;");

gs.print(result);

postMethod.releaseConnection();

Mark my reply as Correct and also hit Like and Helpful if you find my response worthy.

Thanks

Ankur