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Sharing username and password to the next HTML page

I need to share the username and password information to the right next HTML page after succeeding the login. Because the items in the second HTML page will appear according to the user identity and privilege.
I tried sharing the same js file between the 2 HTML pages. The first set the variables and the second get them, but they don't get passed. How do I do it? javascript? jquery? on the server side??
Thanks in advance :)

It can be done in many ways, but as the first language you mentioned is javascript, I will show you in it. So saving data across pages there are variable called session variable and the process of saving/retrieving them is called session management. There are many ways for session management, one most common way is using cookie. You can save the values in cookie, like this:
setCookie("key", "value", expire_time(integer));
And now on next page to get this value you can use:
var val = getCookie("key");
Hope this helps.

Username and password is a sensitive information you have to share it form Server side in these ways:
Use post method to share this information.
Set session on first page For user type and retrieve on very next page.
As your question says second page will appear according to the user identity and privilege. You can set user identity (User Type) and its privileges.

localStorage sign up users how to redirect without losing data

I have a problem with a sign up form. Every time a user creates an account I use localStorage to save the form values. But if after the submit button the user redirects to another page it saves only the last user data who signed up. If after sign up I dont redirect the user to another page I can have more than one users. What can I do to save more users (using localStorage)
the code is:
var passwords=[];
var people= [];
function submitSignUp(){
var usr = signupform.elements["username"].value;
var pass = signupform.elements["password"].value;
people.push(usr);
passwords.push(pass);
localStorage.setItem( 'peoplenames', JSON.stringify(people));
localStorage.setItem('urpasswords',JSON.stringify(passwords));
window.location.href="accountCreated.html";
}
also I use a input type button and not submit because I have the same problem with the submit input. What can I do? Thanks.

Demo
So here is the jsfiddle that should work for you.
some key things are the following.
if(localStorage.getItem('users') != null){
users = JSON.parse(localStorage.getItem('users'));
}
this code checks if the users string has been set for the current page. If users has not been set on local storage is skips this step entirely.
as for processing information from page to page, you will need to do 1 of 3 things.
Send all your calls via ajax. JQuery Get would allow you to stay on the current page while loading content from other pages. You would use the get or post like an IFrame
Navigate through your site using only get methods. Javascript wasn't meant for this, and this doesn't give you much control over the initial state of a webpage.
If you don't want to use Javascript for the rest of your life, or have personal information that shouldn't be held on a local machine. IE: passwords. Use a Server Scripting Language.
In the end it is up to you to decide what to do, but my recommendation would be to use Server Scripting Language and some sort of Database. These are web standards and are marketable skills.

PHP Double-Click Dilemma

We have a problem with users double-clicking on buttons within our application to proceed from screen to screen.
We have implemented the ( onclick="this.disabled=true" ) on our buttons but we are convinced that it is not always sufficient to stop the fast-fingered double-click.
A simple example :-
Screen A has four input fields and a proceed button. When the proceed button is pressed, control is passed to server-side routine to validate info, set some session vars and call screen B.
What appears to happen occasionally is :-
On first click the server-side routine is called and begins validating info and setting session vars. Second-click takes control and again calls the server-side routine and begins validating info and setting session vars -> for us, the session vars are already set and this highlights the problem.
We have looked at tokens but don't think they will solve our problem.
We think that since every PHP application must be vulnerable to this double-click issue there has to be a standard method for resolving it but we have yet to find one.
If you have resolved this issue then we would be grateful if you would like to give us some insights into how we might overcome the problem.
* Thanks for the replies. Loic and Brian Nickel - hard to separate as both going for the token method via timestamp or GUID. We will have to go back and take another look at tokens. After discussion - as a preferred solution for us, we would go with the GUID token concept.

Since double click will basically submit the same form twice you can check the timestamp between two submits.
I'll take the example of stackoverflow because this site is awesome.
Let's say I vote this question up, server side, if my POST request is valid, then my POST request will be treated, and saved.
Then server side, before treating a request, they will check if this same form hasn't been posted in last few seconds (don't they?).
Anyway, my point is, give your forms a name, and when validated, put a timestamp in your users session so you can refuse their post of the same form given a defined amount of time.
Best of luck.

This is a very common problem with a fairly standard solution. Whenever you generate your form, you should generate a unique token like a GUID and stick it in SQL, redis, memcached, the session, or any short term persistent store you have. Stick it in a hidden field. You should be doing one token for each generated form.
When the form gets submitted, atomically check for and remove the token from the store. If it's there the form was submitted for the first time. If not, it's a duplicate.
For bonus points, instead of showing an error on the second submission, you can store the token with the successful result data and use it to render the same success page as you would have if they clicked once.

1) Put a for the eye hidden div (or other element) on z-top of button (opacity:0.01)
2) when once clicked (mousedown) remove div
or:
1) Remove click event when once clicked

How to get cookie from browser, dropped from another application

I have an application URL which generates xyz named cookie for me if the user is logged in, in browser. I want to hit the URL programatically(using Java/JSP) to look if the user is logged in or not.
Now every time when I hit the url from my Java code it doesn't find that cookie (xyz) as it creates new session on every request. This is probably because I am hitting the url from my code. Now how can I come up with this situation so that my application will create cookie in browser and my code will look for that cookie.

The cookie is stored client-side by the browser of the user, so if you call the URL server-side it won't sent the cookie back with the response.

The way you are trying to know that user is logged in or not, can not be achieved through your approach.
Possible workaround for our case is :
Implement the HttpSessionAttributeListener and override the method attributeAdded and attributeRemoved method of it.
When ever session is getting created for logged in user, you would be saving the userName attribute is session most probably. Once the userName attribute is getting saved, attributeAdded method of your listner will get called and you can see the name of user there, once user is logged out, attributeRemoved method will get called by trigger of session invalidate method and your listener will get the notification . You can see the name of user who has logged out.
See below example for detail explanation.
http://www.roseindia.net/servlets/SessionAttributeListenerExample.shtml

How do I preserve radio button and check box value without using cookies after the browser refresh?

I have three radio buttons and 4 check boxes.
I want to preserve the radio button and check box values after the browser refresh.
I do not want to use cookies (due to some other reason).
Could anyone please help me?

I don't think this is possible because HTTP is stateless, cookies or server side scripting provide 'state'.

You could use sessions instead.
EDIT: My bad, I read PHP and not Javascript. However I did find this link after a quick Google search. Session variables without cookies in JS

You might be able to use the hash url.
something like this (don't remember if you need to specify the name of the page as well, but I don't think so):
document.location.href = '#radio1=1&radio2=0'
The hash means it just directs things on the current page and not going to another page (and the browser updates it in the address field, so if the user reloads the page, it will still be there). Then you can read it from javascript as well and set it.
Not as good as using server side sessions, but it is an option :)

If you're using a form to trigger a new page loading you can make the onsubmit event call a javascript function to change the window location and append URL parameters that store the values of the radio buttons. When the page loads you would then read the parameter values from the URL. Something like this:
<script type="text/javascript">
function changeURL(){
var str = window.location+"?radio1=1";
window.open(str);
return false;
}
</script>
...
<FORM onsubmit="changeURL();">
<INPUT TYPE="submit" value="click me" >
</FORM>

A new facility is being developed to allow web sites to store persistent data on the client machine. Available in some browsers already this allows you to save the the radio and checkbox states and recover and restore them next time the user visits your site. For more info see here https://developer.mozilla.org/en/DOM/Storage and here http://dev.w3.org/html5/webstorage

Have some JS on the page submit all radiobutton/checkbox events to the server, and store their state in your database. When the page (re)loads, send this state as part of the HTML.