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How to frame URL with username and password to access Rest API using basic authentication Java?

I have developed Rest API basic authentication, I am able to get the JSON response only when I test using Advanced Rest Client plugin(for Chrome), But how to access the JSON response by passing username and password in URL? How can others consume my rest API? Do I need to give any login popup? How can I achieve this? Any help would be highly appreciated.

your method should be POST and you need to create a encripted token for every login, and you need to active that token up to user logged out or anything else, you can store that token in browser cache, when user logged out then remove that token as well. for every user action you need to verify that token. you can generate token using MD5 also.
i think it may help you.

android jsoup post data to URL

I need to pass data to my URL to get the data.
I have done like this...
document= Jsoup.connect("MYURL").data("PASSCODE", "001100").post();
System.out.println(document);
but I am not getting proper output
Need help. I have check this Links also
This and this also.

Usually login into a web site requires two steps:
You send a get request to get the page, and you extract from there some values like session ID etc, and the cookies.
You send a post request with the values from step 1, and your user name and password.
To know which values you need to send, use your browser in the developer mode (by pressing F12) and examine the traffic.
If you want to write an android app, change the user agent string to match your browser, since some sites send different pages to different clients.
You can see an example HERE.

How to provide a Twitter PIN statically during OAuth authentication

http://twitter4j.org/en/code-examples.html - How to provide a Twitter PIN statically during OAuth authentication, the code example talks about providing a PIN if its already available.
Our application is a standalone java application and would prefer to pass credentials via a property file without any human intervention.

Twitter will provide you the PIN code.
In the example, Twitter4J supposes that somebody will copy the URL it gave (with requestToken.getAuthorizationURL() at line 10) in a web browser and will authorize the application via the browser. After that, the user is supposed to write in the example program the PIN code displayed in its browser.
There are two big drawbacks for you in this example :
Human intervention
Which code will you give to your example program if Twitter do not give you a PIN code (and it will happen if the callback URL of your Twitter application is not "oob") ?
You will have to simulate the human intervention to get the PIN code. Unfortunately, Twitter4J does not seem to get methods to do this (I do not see such method in the Javadoc) so you will have to code yourself the following process :
Retrieve the HTML code of the web page at requestToken.getAuthorizationURL().
This HTML page contains a form whose some parameters will be posted with the username (or email) and the user's password to get the PIN code. These parameters are called "oauth_token" (the temporary OAuth token that you already have), "deny" (a tag used if you do not want to authorize the application) and "authenticity_token". All of them are in <input> HTML tags. Pick them.
Simulate the posting of the form. For this you will have to use the POST oauth/authorize endpoint. The URL is https://api.twitter.com/oauth/authorize?oauth_token=<your OAuth Token>. The request has to be authenticated following the Twitter process for Authenticating requests. This is the content of what you will post : "authenticity_token=<the form's "authenticity_token" parameter>&session[username_or_email]=<user's name or email>&session[password]=<the user's password in clear>". If you don't authorize the app, append "&deny=<the form's "deny" parameter>" in the body message. There will not be any problem for you since you have got the credentials (session[xxx] parameters) in a property file and you picked the other parameters in the form (cf 2.).
The Twitter API will give you back an web page. If the authorization was successful, the PIN will be inside. However, be careful. The PIN takes different names depending on the callback URL. If the callback URL is "oob", the PIN code is called OAuth PIN. It is a 4-digit number somewhere in a <div id="oauth_pin"> HTML tag. Otherwise, it is called OAuth Verifier. It is contained in a URL located in a <div class="happy notice callback"> HTML tag. In this URL, the PIN is the oauth_verifier of the request string. It looks like an OAuth token.
Don't hesitate to have a look at the HTML code of the web pages of the 2nd and the 4th step of my process. It is very useful to understand the process.

I spent some time today trying to figure out how to authenticate without user intervention using Twitter4j Library. I came up with the following:
ConfigurationBuilder conf = new ConfigurationBuilder();
conf.setOAuthConsumerKey(CONSUMER_KEY);
conf.setOAuthConsumerSecret(CONSUMER_KEY_SECRET);
conf.setOAuthAccessToken(TOKEN);
conf.setOAuthAccessTokenSecret(TOKEN_SECRET);
Twitter twitter = new TwitterFactory(conf.build()).getInstance();
By using ConfigurationBuilder, I'm now able to authenticate automatically. Of course you need to set values for CONSUMER AND TOKEN keys accordingly. Here's how I verified it:
User user = twitter.verifyCredentials();
System.out.println("Successfully verified credentials of " + user.getScreenName());
Hope this helps!

FB is sending me a non-URL safe code

FB is sending me a non-URL safe code. While implementing the oauth authentication, FB is sending me a code parameter that is non-url safe.
When i authorize the app i am getting the code parameter with the below value which is not url safe.
AQBQnGZgFNv5EcIB9hZI20GcUgBWru7TE-p_G5NB2HQLX87fyc463aM0BIAiBg_PwRiIuyo0KS1vrDLOLB_SQ0goY31RrNYtZ23lOeaOkGzztarMYH8bJpSYjFOgiXbPhcP6h9YCsizekmcbLpPX_3F7qhsZOLM37UGI7QysvyYOvptpgJ-fm5t-eSViIU_CQH8jTPcQu3BylUVIXS0cEt5T#___=_
When FB is redirecting the control to the my url which is a java web application, i tried to get the code parameter using the code "request.getParameter(CODE)" But i am getting only
AQBQnGZgFNv5EcIB9hZI20GcUgBWru7TE-p_G5NB2HQLX87fyc463aM0BIAiBg_PwRiIuyo0KS1vrDLOLB_SQ0goY31RrNYtZ23lOeaOkGzztarMYH8bJpSYjFOgiXbPhcP6h9YCsizekmcbLpPX_3F7qhsZOLM37UGI7QysvyYOvptpgJ-fm5t-eSViIU_CQH8jTPcQu3BylUVIXS0cEt5T
without #___=_
so i am not able to continue further. But i manually add it to the parameter to fetch the access token it is working fine.
Why is this happening? Please help me.

i think this is signed_request code.

You doesn't need the #_=_ trail to get further, it's not a part of signed_request or code for OAuth2. If you can proceed to further steps it's something other that this.
Once you get code passed to redirect_uri URL you can continue with authentication:
In order to authenticate your app, you must pass the authorization code and your app secret to the Graph API token endpoint - along with the exact same redirect_uri used above - at https://graph.facebook.com/oauth/access_token.
https://graph.facebook.com/oauth/access_token?
client_id=YOUR_APP_ID&redirect_uri=YOUR_URL&
client_secret=YOUR_APP_SECRET&code=THE_CODE_FROM_ABOVE

Why do I get an error 500 when I send POST data to an ASP.NET MVC site through Android?

I'm trying to create an android app to check my tests scores of my engineering school. In order to download the Word containing the scores, I need to login to the portal.
I thought it would be simple to do it by sending a POST request.
After bypassing the problem of the self-signed certificate (or whatever) thanks to the code on this page : Self-signed SSL acceptance on Android
I still get an 500 error while trying to send any POST request to the login page, which is here : https://e-campus.hei.fr/ERP-prod/pc_mv_login.aspx
I tried various codes from the web to send the POST data (especially How to do a HTTP Post in Android? this one). And even on a pure java app, I get a 500.
When I point the URL to another testing page, I manage to get it working, but not on https://e-campus.hei.fr/ERP-prod/pc_mv_login.aspx
Could anyone explain to me why it doesn't work or help me get rid of this error ?
EDIT:
This is what is being sent through my browser (According to chrome developper tools)
__EVENTTARGET:
__EVENTARGUMENT:
__VIEWSTATE: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
Username:******
Password:******
Langues:fr
Button1:Connecter :
This is the string that i send :
String parameters = "__EVENTTARGET=&__EVENTARGUMENT=&__VIEWSTATE="
+ URLEncoder
.encode("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",
"UTF-8") + "&Username="
+ URLEncoder.encode(mUsername, "UTF-8") + "&Password="
+ URLEncoder.encode(mPassword, "UTF-8")
+ "&Langues=fr&Button1="
+ URLEncoder.encode("Connecter :", "UTF-8");

HTTP error 500 just means that the server side code failed. It has a bug, for example a NullPointerException was been thrown over there. If the response body doesn't contain anything sensible (e.g. a stacktrace) so that you could learn how it is caused and so change the request accordingly, then your best bet is to contact the server admin and report about this bug in the server code and ask how to correctly perform a programmatic login.
If that is not an option for some reason, then you should doublecheck if you don't forget to send a specific cookie, header and/or parameter. Probably the server side code was expecting it, but it was null and the code was buggy and hence it totally broke with a 500. I'd suggest to use Firebug to track the entire HTTP traffic and compare it with the headers/parameters you've set. Probably you need to send a specific cookie back? Or you need to send the name=value pair of the submit button? Etcetera.
Update: you're sending the wrong __VIEWSTATE value along. The website runs on ASP.NET MVC which is a component based MVC framework (like as JSF in Java EE). It stores the component tree as "view state". You should not send a random/non-existing/invalidated view state back as paramter, but a valid one. You need to rewrite the HTTP client so that it first fires a GET request on the page with the form and then use a HTML parser (Jsoup?) to extract the value of the hidden __VIEWSTATE input field and finally fire a POST request with exactly that value (and exactly the same cookie in the request header!).
Like as in JSF, the view state is part of CSRF attack prevention. You cannot submit the form without first requesting the form from the website itself in the same session.