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Android java split string via special character fails

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();

Rest Assured code not allowing to use println

I am trying to automate twitter API. when tried to print "js.get("text") using
System.out.println(js.get("text")); I am getting error as
"The method println(boolean) is ambiguous for the type PrintStream"
I downloaded jars and passed in Build path as well "scribejava-apis-2.5.3" and "scribejava-core-4.2.0"
Below code is not allowing me use println for ------>js.get("text")
public class Basicfunc {
String Consumerkeys= "**************";
String Consumersecretkeys="*******************";
String Token="*******************";
String Tokensecret="***************************";
#Test
public void getLatestTweet(){
RestAssured.baseURI = "https://api.twitter.com/1.1/statuses";
Response res = given().auth().oauth(Consumerkeys, Consumersecretkeys, Token, Tokensecret).
queryParam("count","1").
when().get("/home_timeline.json").then().extract().response();
String response = res.asString();
System.out.println(response);
JsonPath js = new JsonPath(response);
System.out.println(js.get("text"));
}
}

Use System.out.println(js.getString("text")); instead of System.out.println(js.get("text"));, because get returns any primitive value.
I think your problem is that your twitter response is actually a list.
Try to use System.out.println(js.getList()[0].get("text")); and be aware that you are only using the first [0] entry and ignoring the rest.

Get relevant part of domain name in Java

If we have an url e.g www.google.de how can I get ONLY the "google"
In Java new URL (url).getHost(); does work but it gives me google.de
and this is not what I want to have.
Thank you
EDIT: If we have something like www.google.co.uk then I also want to have only "google" as result.
I dont want "google.de" or "www.google" I ONLY want "google"

Splitting on a period and selecting the first or second element (whichever is not "www") would work:
URL url = new URL("http://www.host.ext.ext");
String host = url.getHost(); // host = "www.host.ext.ext"
String splitHost = host.split("\\.") // splitHost = { "www", "host", "ext", "ext" }
host = splitHost[0].equals("www") ? splitHost[1] : splitHost[0]; // host = "host"
If there is anything more than http://www. before it, and the extension is potentially more than two "extensions" (.co.uk for instance), then there is no easy way to get just the part you want. As far as I know, you would have to try iterating over a list of extensions and return the part immediately before the longest matching extension.

The most basic solution would be using
System.out.println(url.split("\\.")[1]);
Or you could try this https://stackoverflow.com/a/23079402/2555419
public String getHostName(String url) {
URI uri = new URI(url);
String hostname = uri.getHost();
// to provide faultproof result, check if not null then return only hostname, without www.
if (hostname != null) {
return hostname.startsWith("www.") ? hostname.substring(4) : hostname;
}
return hostname;
}

Java String truncate from URL address

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));

Get last part of url using a regex

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:
".*/([^/#|?]*)(#.*|$)"