regex - Java string split with "." (dot)

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Answer 1

You need to escape the dot if you want to split on a literal dot:

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

Otherwise you are splitting on the regex ., which means "any character".
Note the double backslash needed to create a single backslash in the regex.

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You're getting an ArrayIndexOutOfBoundsException because your input string is just a dot, ie ".", which is an edge case that produces an empty array when split on dot; split(regex) removes all trailing blanks from the result, but since splitting a dot on a dot leaves only two blanks, after trailing blanks are removed you're left with an empty array.

To avoid getting an ArrayIndexOutOfBoundsException for this edge case, use the overloaded version of split(regex, limit), which has a second parameter that is the size limit for the resulting array. When limit is negative, the behaviour of removing trailing blanks from the resulting array is disabled:

".".split("\.", -1) // returns an array of two blanks, ie ["", ""]

ie, when filename is just a dot ".", calling filename.split("\.", -1)[0] will return a blank, but calling filename.split("\.")[0] will throw an ArrayIndexOutOfBoundsException.