The dot "." is a special character in java regex engine, so you have to use "\\." to escape this character:

final String extensionRemoved = filename.split("\\.")[0];

If you want to use split, you need to escape the dot (split expects a regular expression).

List<String> strings = Arrays.asList(myString.split("\\."));

If you only need to remove the last part, you can use replaceAll and a regex:

myString = myString.replaceAll("\\.[^.]*$", "");

Explanation:

• \\. looks for a dot
• [^.]* looks for 0 or more non dot characters
• $ is the end of the string

Try this code it should work:

static HashMap < String, String > newList = new HashMap < > (); 
public static void main(String[] args) throws FileNotFoundException, IOException {

    String inputFile = "input";
    BufferedReader brInput = new BufferedReader(new FileReader(inputFile));
    String line = null;

    while ((line = brInput.readLine()) != null) {

        newList.put(line, "x");
    }

    for (Map.Entry < String, String > entry: newList.entrySet()) {

        String getAfterDot = entry.getKey();
        if (getAfterDot.contains(".")) {
            String[] split = getAfterDot.split("\\.");
            String beforeDot = "";
            beforeDot = getAfterDot.substring(0, getAfterDot.lastIndexOf("."));
            System.out.println(beforeDot);
        }
    }

}

I got this output with above code:

0
0
2
0.3.5
0.2

Try this,

String fileName = "MyFile.xls";     
int dotIndex = fileName.lastIndexOf(".");
String name = fileName.substring(0, dotIndex); // MyFile
String extension = fileName.substring(dotIndex); // .xls
// then you can change and re-construct the file name

You need the regex to only match the last dot, while insuring there are no more dot following it.

To do that you match the dot, which needs escaping, so use \., then use a zero-width positive look-ahead to verify that the rest of the string doesn't have a dot, i.e. use (?=[^.]+$)

Alternatively, use a zero-width negative look-ahead to ensure it cannot find a dot, i.e. (?!.*\.)

I prefer the first, so use \.(?=[^.]+$), which in Java means "\\.(?=[^.]+$)"

When you say split you are using delim as a regex pattern. It is treated differently. Please have a look to this regular expression.

But when you are using delim in sysout you are using it as string. the difference is obviuos

It might be easier to just assume that files which end with a dot followed by alphanumeric characters have extensions.

int p=filePath.lastIndexOf(".");
String e=filePath.substring(p+1);
if( p==-1 || !e.matches("\\w+") ){/* file has no extension */}
else{ /* file has extension e */ }

See the Java docs for regular expression patterns. Remember to escape the backslash because the pattern string needs the backslash.

^{I see only solutions here but no full explanation of the problem so I decided to post this answer}

Problem

You need to know few things about text.split(delim). split method:

1. accepts as argument regular expression (regex) which describes delimiter on which we want to split,
2. if delim exists at end of text like in a,b,c,, (where delimiter is ,) split at first will create array like ["a" "b" "c" "" ""] but since in most cases we don't really need these trailing empty strings it also removes them automatically for us. So it creates another array without these trailing empty strings and returns it.

You also need to know that dot . is special character in regex. It represents any character (except line separators but this can be changed with Pattern.DOTALL flag).

So for string like "abc" if we split on "." split method will

1. create array like ["" "" "" ""],
2. but since this array contains only empty strings and they all are trailing they will be removed (like shown in previous second point)

which means we will get as result empty array [] (with no elements, not even empty string), so we can't use fn[0] because there is no index 0.

Solution

To solve this problem you simply need to create regex which will represents dot. To do so we need to escape that .. There are few ways to do it, but simplest is probably by using \ (which in String needs to be written as "\\" because \ is also special there and requires another \ to be escaped).

So solution to your problem may look like

String[] fn = filename.split("\\.");

Bonus

You can also use other ways to escape that dot like

• using character class split("[.]")
• wrapping it in quote split("\\Q.\\E")
• using proper Pattern instance with Pattern.LITERAL flag
• or simply use split(Pattern.quote(".")) and let regex do escaping for you.

In one line with substring and indexOf:

String output = str.substring(0,str.indexOf(".",str.indexOf(".")+1));

You can solve this with regex (given you only need a group of word characters between the last "/" and "."):

    String str="core/pages/viewemployee.jsff";
    str=str.replaceFirst(".*/(\\w+).*","$1");
    System.out.println(str); //prints viewemployee

The documentation on split() says:

Splits this string around matches of the given regular expression.

(Emphasis mine.)

A dot is a special character in regular expression syntax. Use Pattern.quote() on the parameter to split() if you want the split to be on a literal string pattern:

String[] words = temp.split(Pattern.quote("."));