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Skip to a Page on Google Forms

I'm programatically creating a Google Form using Google App Scripts right now, and my script automatically generates a page for each of the respondants that the form will be emailed out to.
Currently, the form has a dropdown at the beginning that the person uses to select their page, and then it forwards them to it, however, I'd prefer it if the email already sent out a pre-responded form that already started them on their page.
The reason I'm throwing them all into a single form is because I'm using Triggers to detect the form's submission, and since I have more than 20 people that require a form, that would break the trigger limit.
EDIT:
Example code:
the Form.getPublishedUrl() function returns the URL to respond the form, starting on the first page, however, I would like a URL to respond to the form starting on the 3rd or 4th page.
While the FormResponse.toPrefilledUrl() will allow me to autofill the correct response on the first page, it will not allow me to start the responder on the page that he would be forwarded to based on his pre-filled response on the first page.
tldr; :
How do I get a link to a specific page on a Google Form using Google App Scripts?

This should be a comment but don't have 50 rep yet...
Why do you need various HTML pages?
I think you should refactor your code so every form is on the same HTML and everything shows and hides with Jquery or Javascript. You can send the same URL but with differente URL parameters.
If this isn't your case, you should post some code of yours and detail a little bit more your question. Also, wich limit does it triggers?

Redirect using post and get the succes of the redirect

Say, I have a simple form on my website having three fields : name, password and email.
I have to get these information from the users, and keep in my database.
Then redirect to another website and send all these information using post.
I also have to know whether the user was successfully redirected to that site(HTTP STATUS 200).
Here's how I'm doing it:
For Point 1, I'm simply submitting the form.
After the data has been successfully saved in my database, I'm rendering the following form with hidden fields. This gets submitted and user gets redirected to anotherwebsite.com
<form id="form_id" action="https://www.anotherwebsite.com/form" method="POST">
<input type ="hidden" name ="name" value ="$name">
<input type ="hidden" name ="password" value ="$password">
<input type ="hidden" name ="email" value ="$email">
</form>
<script> document.getElementById('form_id').submit(); </script>
Problems:
I don't think my strategy to achieve point 1 and 2 is correct. I need a better solution. Submitting the form, then rendering a page with hidden fields and submitting it again to redirect to another site just doesn't feel right.
I have no clue to achieve the 3rd point.

Based on your question you might try this approach:
create a form with name, password, email fields in a file ( HTML ).
Submit the form to server.
On the server side get the data (including the form attribute in a variable) and save it to database.
then redirect to the given website ( using the variable you've stored in step 3 ).
You can easily know the status ( 202 or any error) using any of server side scripting language.

If you are sending the user to another website, the only way to know that the user was successfully redirected is for that website to notify you in some manner. Once the user leaves your page (and that's what a redirect is - it tells the browser "leave this URI and go to this URI instead"), the scripts on that page stop running, so they can't collect any further information.
If you just need to know that the information was submitted successfully, your script could POST the data in the background, wait for a 200 response, then redirect after the information has been submitted. But that may not meet your requirements, since you still won't know if the redirect succeeded.
Another possibility which does allow you to know whether the page on the other site loaded correctly would be to open it in a new browser window/tab instead of redirecting. This is the only way to keep your page active (and, thus, your scripts able to run) while loading another page. However, it introduces other issues, like what to do with the original page. (Leave it open in the background (likely to confuse the user) or close itself after seeing that the new URI has loaded (could cause undesirable visual artifacts as one window/tab opens and then the original one closes; destroys browser history)?)
If at all possible, having the final destination site notify you when the transaction completes is almost certainly the best way to go.

To achieve point 3 you need to use cookies if you are actually trying to implement a login-cum-membersarea system. Othewise, you simple need a redirect inside a condition statement.
my $cgi = CGI->new;
if (condition) { print $cgi->redirect('https://www.examplesite.com/file.html') }
for a general way of doing point 1-2, you can look at the tutorial here:
http://practicalperl5.blogspot.com/

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

Shall I use javascript for page submission?

I am working on a big site, and in the site there is a search module. Searching is done by using a a lot of user submitted values, so in pagination I must pass all these data to the next page, appending the values to url make the url very big.
Sso how can I solve this issue? I am planning to use a javascript based page submission (POST) with all the values in hidden fields to the next page the read all the values from the next page.
Will it cause any problems? Or should I use database to keep the search criterias?

I would create a server side object, possibly with a database backend which is updated by the different pages.
It is at my opinion the most clear and easy solution. Giving parameters from page to page, either by post or javascript or cookie will work too but it's more of a quirk in my experience.
Also if a search query is so complex that it needs multiple pages to create it, it might be helpfull for the user to have all the data stored on the server so he can change it more easily by switching back and forth between the different pages.

I would store all the search criterias in some kind of session-store on the server when the initial search is being triggered.
For pagination I would retrieve the criterias from the session-store and then just show the appropriate results. Also I would append some kind of key to the pagination links (so this would be the only hidden post-field) under which the search criterieas can be found.
Even though the session is per user, you might have several search windows open within the same session, and you don't want to mess them up with the pagination.

In order to make a reliable search with pagination, we need to do a bit more than normal.
We need to handle the following cases.
Once search is done, user may choose to do browser back and forward. Here, if you are doing form submission on every page, it would be an overload. Also, if user presses browser refresh button, it will unnecessarily warn him that data is being submitted.
Searching on a large database with lots of criteria is costly. Hence, optimization is important.
So you should NOT do the following:
Submit data on every page change
Not store data in cookie. (This is not secure and not even reliable.)
For large database with complex query, cache the result in session.
In case, you need very up-to-date and real-time result, ignore point (3) and try doing partial search for every page.
Thus, for your case, you can do the following:
When user searches first time, make the form POST data to a search page.
This search page will store the search query in session and generate a unique id for it.
Now render the result page. The result page will be passed the search id (generated in point 2) and the page number. Example result.aspx?searchId=5372947645&page=2
The result page will puck up the query from session using the searchId and then provide result based on the page number sent.

Using hidden fields and POST method should be fine too unless you are able to get them on the next page right.

To supplement Sarfraz's answer...
It's not necessary to use Javascript to make a POST.
<form action="destination_url" method="POST">
...
</form>

Redirect to previous view and inject a javascript alert

I need to do the following:
I have a textbox, which appears in every page of the site, that allows to subscribe to a newsletter. This I've done already and the user is redirected to previous view after subscription.
I'd like to add a javascript alert to the page the user is returned to, something like "Thanks for subscribing". How can this be done?
Thanks in advance.
EDIT: Propably it's not clear from the post tags. I'm using ASP.NET MVC 2 Preview 1

If you are doing a HTTP redirect, then the page you will render needs to be passed some information so that it knows to include the javascript to open the alert box (adding an optional element to the page might be a nicer way to do this).
That information needs to be stored either in a browser cookie, or in a session store (which is keyed from a browser cookie). You can remove this once you've rendered your message, so that it is only shown the first time you visit that page after the redirect.