Get Firebase Dynamic Link Data from String Input url Flutter - flutter

Dynamic links work great for 98% of our users. However, there are still a group of users which have difficulty with them or do not know how to use them.
I want to add a feature which would let users paste their link into the app, and then we extract the data from the link and handle it normally. This will also serve as a backup for when the links are down or misbehaving. It will also allow our customer service team to get data from a link when customers share them with us.
The problem is, there doesn't seem to be a way to manually pass in a dynamic link to retrieve the dynamic data.
Does anyone know how this can be achieved?

Here is my attempt at your question.
I am assuming what you mean by the dynamic data is the underlying deeplink along with the parameters associated with the deeplink.
void dynamicLinkToDeepLink(String dynamicLinkString) async {
final details = await FirebaseDynamicLinks.instance.getDynamicLink(Uri.parse(dynamicLinkString));
// your deep link can be accessed as follows
details!.link;
}
You have to safeguard the above code as you see fits when you use it. You will have to wrap FirebaseDynamicLinks.instance..... with a try catch block and you will also have to check if the value of the returned link is not null before acccessing details!.link

Related

Redirect URL using Firebase Dynamic / Deep Links is losing query parameters

In my Flutter (Android/iOS) app I am using Firebase Dynamic Links for Patreon.com OAuth2 apis.
My dynamic link is https://myappname.page.link/patreon
The deep link is https://myappname.net/patreon
Patreon is using the https://myappname.page.link/patreon as a redirect_url , and is supposed to append some parameters to it, so it looks like
https://myappname.net/patreon?code=xxx
However, all I receive inside my app is the naked url https://myappname.net/patreon
There are no parameters attached to it.
So how can I tell Firebase to preserve the query parameters Patreon is attaching to the redirect_url?
As an alternate question, is there a better way to listen for incoming response inside of a Flutter app, without the use of Dynamic Links?
You loose all parameters by using that.
If you're relying on Patreon to send back that parameter I'd suggest to generate a small proxy where you can redirect your calls to the dynamic link by generating it on the fly.
So:
Patreon shares www.myhost.com/supah-link?p1=aaa&p2=bbb
Your micro-service which runs on www.myhost.com/supah-link receives the call
You generate a dynamic link like the following:
https://example.page.link/?link=https://www.example.com/someresource&apn=com.example.android&amv=3&ibi=com.example.ios&isi=1234567&ius=exampleapp&p1=aaa&p2=bbb
NOTE: Pay attention to the &p1=aaa&p2=bbb parameters added
Return a 302 and redirect to your newly generated link
Based on how you configure it from the console this link can redirect to the store or your app, in your app you can listen for the link as follows:
FirebaseDynamicLinks.instance.onLink(
onSuccess: (dynamicLink) async => handleDeepLink(dynamicLink?.link),
);
In handleDeepLink you can parse your custom query parameters.
NOTE: The URL parameter you pass via the dynamic link HAS TO BE ENCODED! Which means your link will look more like this:
https://example.page.link/?link=https%3A%2F%2Fwww.example.com%2Fsomeresource%26apn%3Dcom.example.android%26amv%3D3%26ibi%3Dcom.example.ios%26isi%3D1234567%26ius%3Dexampleapp%26p1%3Daaa%26p2%3Dbbb

Use Login Robox using username only like on reward sites

Hi i know this is probably not the place to ask this but i m stumped at the moment as i cant seem to find any reference or docs relating to working with Roblox. I mean sure they have an auth route etc but nothing detailed. I want to login user using username and give them roblox based on different actions they take on the site like completing surveys etc. Can anyone please give me links to some resources that would come in handy for the particular purpose. Thank you.
Roblox does not support any OAuth systems, but you still can use HttpService:GetAsync() function to get strings/data from web site(if the page in website display that text), the way to keep data that you recieved from url(web page) safe is to store script with HttpService:GetAsync() function in server side(example: RobloxScriptService). You need to allow http requests in your GameSettings -> Security in roblox studio. Script example:
local HttpService = game:GetService("HttpService")
local stringg = HttpService:GetAsync("https://pastebin.com/raw/k7S6Ln9R")
print(string)
--Should outpud data written ot the web page, you can use any web page to store data even your own
The only two things that left is to make your web server rewrite the page, or just use some databases at your web site by placing their url into loadstring() function.
Now you just need to parse the string given by url to use it's data.
The pastebin url that i wrote into loadstring() just an example, you can write whatever you wan, but again you need to parse the data that you got from url, or just convert the string into type of text like on the page, and then just check is they written at url/webpage. Example:
local writtenpass = game.Players["anyplayer"].PlayerGui.TestGui.Frame.PasswordTextBox.text
local writtenlogin = game.Players["anyplayer"].PlayerGui.TestGui.Frame.LoginTextBox.text
local HttpService = game:GetService("HttpService")
local response = HttpService:GetAsync("https://pastebin.com/raw/k7S6Ln9R")
local istrue = string.find(response, "{ login = ".. writtenlogin .." pass = ".. writtenpass .." }")
print(istrue)
if istrue == 1 then
print("exist!")
--whatewer actions if login and pass exist
end
You can wiew the page here https://pastebin.com/raw/k7S6Ln9R
Well that a lot of damage!
If it helps mark me

How to remove query string from Firebase Storage download URL

Problem:
I need to be able to remove all link decoration from the download URL that is generated for images in Firebase Storage.
However, when all link decoration is stripped away, the resulting link currently would return a JSON document of the image's metadata.
Context:
The flow goes as follows:
An image is uploaded to Firebase from an iOS app. Once that is done the download URL is then sent in a POST request to an external server.
The server that the URL is being sent to doesn't accept link decoration when submitting image URLs.
Goal:
Alter the Firebase Storage download URL such as it is stripped of all link decoration like so:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
Notes:
The problem is twofold really, first the link needs to be manipulated to remove all the link decoration. Then the behavior of the link needs to changed, since in order to return an image, you need ?alt=media following the file extension, in this case .jpg. Currently, without link decoration, using the link with my desired structure would return a JSON document of the metadata.
The current link structure is as follows:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg?alt=media&token=[TOKEN]
Desired link structure:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
The token is necessary for accessing the image depending security rules in place, but can be ignored with the proper read permissions. I can adjust the rules as needed, but I still need to be able to remove the ?alt=media and still return an image.
Building up on Frank's answer, if you access to your associated Google Cloud Platform project, find the bucket in the Storage tab and make this bucket public, you will be able to get the image from here with the format you wish. That is, you will not be accessing through Firebase
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
but through Google Cloud Storage, with a link like
https://storage.googleapis.com/[bucket_name]/[path_to_image]
Once in your GCP project Console, access the Storage bucket with the same name as the one you have in your Firebase project. They are the same bucket. Then make the bucket public by following these steps. After that, you will be able to construct your links as mentioned above and they will be accessible with no token and no alt=media param. If you do not want to make the public to everyone, you will be able to play around with the permissions there as you wish.
You could split the url string into two halves by using String.componentsSeparatedByString(_ separator:)
Storage.storage().reference().child(filePath).downloadURL(completion: { (url, error) in
let urlString = url.absoluteString
let urlStringWithoutQueryString = urlString.componentsSeparatedByString("?").first!
})
Calling .downloadURL on a StorageReference will return you that URL, but this method can be used to remove the query string from any URL. String.componentsSeparatedByString(_ separator:) breaks a String into an array of Strings, splitting the string by any occurrence of a given separator, in this case ?.
NOTE this method assumes that ? occurs only once within the url string, which I believe is the case for all Firebase Storage urls.
You should treat the download URL that you get back from Firebase as an opaque string. There's no way to strip the parameters from a download URL without breaking that download URL.
If you want to allow public access to the files in your bucket with simpler URLs, consider making the object in your (or even your entire) bucket public.

OpenGraph: how can i specify a filter in FB.api?

I have built a Facebook app using OpenGraph that permits the users to write reviews on concerts, so that I've defined a concert_id attribute on which the user can insert a review.
Now I would like to show all the reviews inserted for a certain concert_id but cannot find a way. If I do (in JS)
FB.api('/me/MY_APP:action', { limit: 0}, function(response) {
console.log(response);
});
I get all items. This app has to be consumed by mobile, I think it is bad to get all items and, then, filtering only the concert_id i need. What do I have to do to apply a where condition in OpenGraph to a custom action?
As far as I can tell from the API and the Facebook developer pages, it's not possible to filter a call by custom action property using the public Open Graph API.
Two options I can think of:
Option 1:
Implement the category filter by creating custom category objects:
if "review" is a custom action and
GET https://graph.facebook.com/me/[name_space]:review
returns all review actions then
GET https://graph.facebook.com/me/[name_space]:review/scifi_movie
GET https://graph.facebook.com/me/[name_space]:review/action_movie
return actions specific to movie type, where scifi_movie and action_movie are custom objects. You would need to create one object type for each category.
Option 2:
Implement a custom action for each category, e.g.
review_scifi_movie
review_action_movie
These are not particularly elegant solutions but perhaps useful as a hack if nothing else works and you really don't want to do filtering on client side.
The Facebook API will not return individual published objects for a particular action, but that's not your only problem. By the look of it, you're trying to bring in ALL the reviews given for a concert, right? (Meaning those by other users too).
The "/me/" part of the Facebook API call will only return those published actions made by the user that is currently logged in. That won't work for you, as you want those of all your users
The only suggestion I can give is to create a simple web service, where you store all the reviews given for the various concerts. Use this service to pull in reviews given for a particular concert. (I use a similar methodology for reviews in an app of my own).
I dont understand javascript or opengraph..
But when I required in JAVA to fetch reviews made by any user I have used FQL for that and It retrived me all the reviews and FQL also used to fetch all the tables related to Facebook.
I don't think that you can pull that off with the JS SDK.
You can do that in your server though, and since this is a mobile app (or has a mobile version) then that's another good reason to remove this from the client responsibility.
In the server side you can ask facebook for the published actions as you posted, filter them and then return the response.
Another thing that you can do is to save each published action in your db (on each action post you should get an id back from facebook, just persist that) and then you can easily filter the published actions according to what ever criteria you want/need (since you are no longer restricted by the facebook api).
The open graph thing is still pretty new and not tat mature, for example you can't use FQL with it, something that could have been handy for your case.
Regardless though I think that a server solution is best for calculations when mobile is concerned.
i don't know exactly but try this
if (session.authResponse) {
FB.api('/me', {
fields: 'name, picture' // here mention your fields
},
function(response) {
if (!response.error) {
//here response value
});

Simplest example for sending post data via links in Zend Framework

Starting with Zend and I´d like to know what is the simplest way of sending POST data to another page, not by forms, but by some link in my view instead. Thanks :)
You can't send POST data through a link. At least not through a normal link. Link can only carry GET data.
If you need to send POST over a link it's most certainly a design flaw.
If you're 100% sure, that you need it, you can do that using jQuery and onclick event. It`s not possible to do it without javascript. Other option would be to send it using form with hidden fields with single submit button visible - that would even work without javascript.
Normal hyperlinks in HTML are sent with GET requests and are not supposed to change the state of the resource being accessed. This is known as being idempotent. You can repeat the request over and over, and the result of each succeeding request to the same URL is the same as the first one.
POST requests don't have this restriction and are intended for when the user needs to change something (such as creating a new resource.)
It's not possible to send a POST request via a normal HTML link. And even if you find a way, it breaks an almost universal expectation that web users have. What are you trying to accomplish? Maybe there's a better way.
But to answer your question, you could use something like jQuery to capture the "click" event and make it do a POST request:
$('.my-link').click(function() {
var url = $(this).attr('href');
var data = {};
$.post(url, data, function() {
window.alert('success!');
});
return false;
});
If your URL has any query parameters, i.e. "?foo=bar&baz=bum", then you'd probably need to strip them off of the URL and pass them as a second parameter to the $.post() function. This is left as an exercise for the reader. ;-)