I have a deeplink that contains an encrypted hash id param, see below URL-
which redirects to login page if users are not already logged in with initial request in ForwardURI param, see below URL-
The issue is when Spring binds forwardURI into LoginForm, it trims off the equals padding from encrypted hash and that results in a corrupt id which we can’t decrypt.
Is there a way to tell Spring not to trim these reserved characters in params?
In my akka http app I receive a lot of errors like this one:
Illegal request-target: Invalid input 'v', expected HEXDIG (line 1, column 72)
It happens when I simply add a '%' character to the URL. In this case akka exoects a hex value.
Is it possible to somehow 'clean up' or encode/ignore such characters?
This question already has an answer here:
When i decompile my apk i saw that anyone can see the url I'm connecting to.The api key that i'm using is also easily accessible.So is there any way to encrypt the String url so when someone try to sneak it there won't be any proper string or api mentioned?
Having a problem with HttpWebRequest decoding my encoded URL.
var requestUrl = "https://www.google.com/webmasters/tools/feeds/http%3A%2F%2Fwww%2example%2Ecom%2F/crawlissues/"; var request = (HttpWebRequest)WebRequest.Create(requestUrl);
When looking at end request URL is becomes:
Which of course returns a 400 Bad request. I am guessing it is something todo with the URI class rather than HttpWebRequest. How do I stop this from happening?
This is my url .my problem is address not appending to urlpath . plz help me
var address:stringaddress = "American Tourister, Abids Road, Bogulkunta, Hyderabad, Andhra Pradesh, India"let urlpath = NSString(format: "http://maps.googleapis.com/maps/api/geocode/json?address="+"\(address)")