I am very new to this concept ,when i share my image and text to other app,my domain name was visible but i want to encrypt it and when user click that link it should be redirect to my play store app URL ,i google it but I am getting any solution for ,i saw this functionality in many app please help me how to d this here below one of the app show like this i need to encrypt and redirect to play store without opening that URL i need to redirect to play store please any one help me how to do that,here below one of the app using like this encryption and redirection with out showing their Main URL .
Here below the Image
What you see in the image is not encrypted. It's an url Shortener. And in this case it's from google. Documentation about the Google URL Shortener
I think to make safe calls you need to encrypt it while sending and when user click on that image now you want to redirect to play store URL, so now decrypt it.
I found one library to do this job below is the link for example
https://github.com/simbiose/Encryption/blob/master/Examples/Android/app/src/main/java/se/simbio/encryption/MainActivity.java
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I am making a app where user can upload any image and get a link for that image. By using that link user can share that image something like imgur site. For this purpose I am using firebase storage. Now the thing is that link is containing some information and now I'm worrying the user might exploit things.
https://firebasestorage.googleapis.com/v0/b/"myappname".appspot.com/o/Image.png?alt=media&token="some token number"
So my question is, Is it possible for anyone to exploit or do any harmful thing through above link?. I'm not using any authentication rules for storage.
What you're showing is a download URL to a file in Cloud Storage. Anyone that has this URL can read the file associated with it, regardless of the security rules you've set up.
That is precisely the level of abuse you're opening yourself up for when you share this URL with someone: they can read the file, and/or share the URL with others who can then do the same.
This is the same as when you'd give them the URL to another file that is hosted on a service you own - and which is not secured.
I just created a short dynamic link for this url https://m.example.com/invitation?gameid=1234&referrer=555.
It is working fine, but when i click the shortlink in browser it is directing to my page but it is showing all the sensitive information like gameId and referrer in browser address bar. I don't want to show that.
Is there any way to show short link even when i click the shortlink in browser also or anyway to hide the sensitive information.
Firebase Dynamic Links do not provide mechanism to hide link data from person who have access to the link. In addition to exposing the link content in browser address bar (as you already found out), any person can get link details by going to link debug page https://firebase.google.com/docs/dynamic-links/debug . Append ?d=1 or &d=1 to the link to get here.
If you need to hide information, you can consider encrypting the information, like https://m.example.com/invitation?encrypted_data=<base64 data here>
Another option is to store data in database and having only data identifier in your link, like https://m.example.com/invitation?data_id=<identifier>
If you think your use case is important or widely used, feel free to open Firebase support ticket. We always open to feature requests and suggestions how to improve our product.
I am using the Pinterest Android SDK to authenticate and read pins from Boards I follow.
The issue is in my app, I need to to open the URL of the pin in a webview to able to read the pin contents in my app itself without leaving my app.
Currently, I can use the "url" field from the getPin() API response json, but the URL redirects from the webview to the pinterest app (leaving my app).
Is there a way to do this?
Thanks.
do you intend to show the entire pin or just the image? If you use the URL of the getPin() API then it is the URL to a "Pin" which is a Pinterest object and will always have a "Pinterest.com". Hence it will redirect itself to Pinterest app, which is the correct behaviour.
However, you can get the URL of the image from the Pin that you retrieved using getPin and display that within the app.
The other problem with your approach is that it tries to replicate the behaviour of Pinterest itself which is in violation of Pinterest policies and they may ban your app. So my suggestion would be to use the URL of the image of the pin and display inside an imageview of your app using Picasso or Glide.
All the best...
I have an appengine connected android project. It's using endpoints. Anyway, I have some images in the blobstore. The android app has the url to each image. Normally, android needs a full image. But for my GridView I want to display a number of thumbnails. I know how to fetch images from a server (here the blobstore). My question is, is there a way to ask the blobstore to send me a thumbnail of an image? The call needs to come from the android side.
You could use get_serving_url()
You can use it to generate a link to the image in the blobstore and can specify the size of its "longest dimension ... preserving the original aspect ratio":
images.get_serving_url(your_blob_key,size=300)
This will produce a URL like this: http://lh6.ggpht.com/iQLnaVoMEhOWzMummf_PzD8frysTBIP6FsTCYpEuf0FA8KaK72T0zreND_pi4vZyUJQvv72x_rXLisIJOiXbb2IWzBc=s300
As you can see, setting size=300 simply adds =s300 to the end of the URL, so you could skip adding this and just add =s300 to the url in your app.
Be aware that this creates a "Public but not guessable URL". This essentially means that anyone could potentially access the link generated if they have the URL.
The other issue with this is that you would need to generate the link from within App Engine. However if you NEED to be able to just access it directly through the app you could generate the link during upload and store this in the datastore along with the blobstore key. You could then send this link to the app, in-place of the existing link being sent.
I am trying to create an application in which will be notifying the user about the phishy websites.So for this i need to get the url which the user has opened in the default browser.
After a lot of google search and stackoverflow search i am not getting any idea how to develop the app.Can you guys help me ...how i could get the url entered by the user (just want the url accessed).
Or any other way round to develop this type of app in android.
Any help would be highly appreciated.
Maybe you can proceed the following way.You can read browser bookmarks.Then see the arrange pattern of the bookmarks.Maybe it is sorted according to the accessed time.So you can get what you need