We are getting url from JSON Response and which we open in in Chrome.The page loads , there is submit button which we click then it redirect to url as :-
https://www.google.com/AB1234
We need the need to retrieve only "AB1234" value from url.
tried following code to get value ="AB1234"
String url = driver.getCurrentUrl();
int index=url.lastIndexOf("/");
String result = url.substring(0,index);
but here getting initial part of url:https://www.google.com/
You need to call substring function with index +1 .
Try below code :
String url = driver.getCurrentUrl();
int index = url.lastIndexOf("/");
String result = url.substring(index + 1);
To parse a URI, it's likely a good idea to use a URI parser.
Given http://example.com/bar
String path = URI.create(driver.getCurrentUrl()).getPath();
will get you '/bar'.
Given http://example.com/bar/mumble the same code gets '/bar/mumble'. It's unclear from your question whether this is what you want. Nevertheless, you should at least start the parse as above.
Related
I would like to attach a platform parameter to a url with ? if the url has no query string and using & if url has a query string
SO i have added the following
String api_url;
//costructor next to assign apiurl value
//method to extract url and process request
processData(){
String apiUrl = "";
String[] urlParams = this.api_url.split("\\?");
if (urlParams.length > 0){
apiUrl = this.api_url+"&platform="+tokenService.getToken(AppDetailsHelpers.AppSettingsKeys.PLATFORM);
}else {
apiUrl = this.api_url+"?platform="+tokenService.getToken(AppDetailsHelpers.AppSettingsKeys.PLATFORM);
}
}
The above always evaluates the urlParams to a an array even when a url doesnt contain the ?
Example for a url
http://test.com
is resolved with the above code as
http://test.com&platform=12
But i expected it to be as http://test.com?platform=12
I have tried adding
String[] urlParams = this.api_url.split("?");
But it throws an error of Dangling metacharacter. What am i missing out on this. Why does this fail.
This is expected behaviour for String#split. Running "http://test.com".split("\\?") returns an array with one element, "http://test.com". So, just update your condition to if(uriParams.length > 1).
You could also consider parsing your String to a Uri, as you may not need this check and could possibly instead use:
Uri.parse(api_url)
.buildUpon()
.appendQuery("platform", tokenService.getToken(AppSettingsKeys.PLATFORM))
.build().toString();
I am working on "Forgot Password". I am trying to create a reset token with email + current_time. email is user login whilst code will check if time >= 5 minutes then this link will not work. Here is my code:
// preparing token email + time
Date now = new Date();
String prepareToken = "?email="+email+"&tokenTime="+now.getTime();
// encrypt prepareToken value
Encryptor enc = new Encryptor();
resetToken = enc.encrypt(resetToken);
The token will be sent as for example as http://domainname.com/ForgotPassword?resetToken=adj23498ljj238809802340823
Problem:
When user click it then I got as request parameter and obviously decrypt this parameter but how can I get email in one String + time as another String
Please advise
If your issue is simply parsing the decoded String to get some sort of Map of your parameters, I'd suggest you to read Parse a URI String into Name-Value Collection .
Hope it helps.
EDIT :
Assuming you have the splitQuery(URL url) method from the previous link and that you successfully decoded the token :
public String getEmailFromToken(String decodedToken) {
// if you decoded your token it will looks like the prepareToken String
String stubUrl = "http://localhost"+decodedToken;
Map<String,String> map = splitQuery(new URL(stubUrl));
return map.get("timeToken");
}
I created a properly formed URL to respect the URL syntax.
With little tweak, you should be able to implement splitQuery for a String. I hope you can manage that.
I have an URL address like: http://myfile.com/File1/beauty.png
I have to remove http://site address/ from main string
That mean result should be File1/beauty.png
Note: site address might be anything(e.g some.com, some.org)
See here: http://docs.oracle.com/javase/tutorial/networking/urls/urlInfo.html
Just create a URL object out of your string and use URL.getPath() like this:
String s = new URL("http://myfile.com/File1/beauty.png").getPath();
If you don't need the slash at the beginning, you can remove it via s.substring(1, s.length());
Edit, according to comment:
If you are not allowed to use URL, this would be your best bet: Extract main domain name from a given url
See the accepted answer. Basically you have to get a TLD list, find the domain and substract everything till the domain names' end.
If, as you say, you only want to use the standard String methods then this should do it.
public static String getPath(String url){
if(url.contains("://")){
url = url.substring(url.indexOf("://")+3);
url = url.substring(url.indexOf("/") + 1);
} else {
url = url.substring(url.indexOf("/")+1);
}
return url;
}
If the url contains :// then we know that the string you are looking for will come after the third /. Otherwise, it should come after the first. If we do the following;
System.out.println(getPath("http://myfile.com/File1/beauty.png"));
System.out.println(getPath("https://myfile.com/File1/beauty.png"));
System.out.println(getPath("www1.myfile.com/File1/beauty.png"));
System.out.println(getPath("myfile.co.uk/File1/beauty.png"));;
The output is;
File1/beauty.png
File1/beauty.png
File1/beauty.png
File1/beauty.png
You can use the below approach to fetch the required data.
String url = "http://myfile.org/File1/beauty.png";
URL u = new URL(url);
String[] arr = url.split(u.getAuthority());
System.out.println(arr[1]);
Output - /File1/beauty.png
String s = "http://www.freegreatpicture.com/files/146/26189-abstract-color-background.jpg";
s = s.substring(s.indexOf("/", str.indexOf("/") + 1));
How do I get the last part of the a URL using a regex, here is my URL, I want the segmeent between the last forward slash and the #
http://mycompany.com/test/id/1234#this
So I only want to get 1234.
I have the following but is not removing the '#this'
".*/(.*)(#|$)",
I need this while indexing data so don't want to use the URL class.
Just use URI:
final URI uri = URI.create(yourInput);
final String path = uri.getPath();
path.substring(path.lastIndexOf('/') + 1); // will return what you want
Will also take care of URIs with query strings etc. In any event, when having to extract any part from a URL (which is a URI), using a regex is not what you want: URI can handle it all for you, at a much lower cost -- since it has a dedicated parser.
Demo code using, in addition, Guava's Optional to detect the case where the URI has no path component:
public static void main(final String... args) {
final String url = "http://mycompany.com/test/id/1234#this";
final URI uri = URI.create(url);
final String path = Optional.fromNullable(uri.getPath()).or("/");
System.out.println(path.substring(path.lastIndexOf('/') + 1));
}
how about:
".*/([^/#]*)(#.*|$)"
Addition to what #jtahlborn answer to include query string:
".*/([^/#|?]*)(#.*|$)"
I have a String holding a URL in this format: http://hello.world.com/service/sps/f4c0e810456t
And I would like to extract the last part of the URL, i.e. f4c0e810456t.
I can do it with substrings:
System.out.println(s.substring(s.lastIndexOf("/") + 1, s.length()));
Or regexp however looking for something more elegant using URL/URI objects but couldn't find something.
Any ideas...?
If you can change the URL to "http://hello.world.com/service/sps/?f4c0e810456t" then you could use the getQuery() method (both on URL and URI).
Example with URL and split (it wrap regular expression for you):
String address = "http://hello.world.com/service/sps/f4c0e810456t";
URL url = new URL(address);
String [] str = url.getPath().split("/");
String result = str[str.length-1];