How do I get the first n characters of a string without checking the size or going out of bounds?

Solution 1:

Here's a neat solution:

String upToNCharacters = s.substring(0, Math.min(s.length(), n));

^{Opinion: while this solution is "neat", I think it is actually less readable than a solution that uses if / else in the obvious way. If the reader hasn't seen this trick, he/she has to think harder to understand the code. IMO, the code's meaning is more obvious in the if / else version. For a cleaner / more readable solution, see @paxdiablo's answer.}

Solution 2:

Don't reinvent the wheel...:

org.apache.commons.lang.StringUtils.substring(String s, int start, int len)

Javadoc says:

StringUtils.substring(null, *, *)    = null
StringUtils.substring("", * ,  *)    = "";
StringUtils.substring("abc", 0, 2)   = "ab"
StringUtils.substring("abc", 2, 0)   = ""
StringUtils.substring("abc", 2, 4)   = "c"
StringUtils.substring("abc", 4, 6)   = ""
StringUtils.substring("abc", 2, 2)   = ""
StringUtils.substring("abc", -2, -1) = "b"
StringUtils.substring("abc", -4, 2)  = "ab"

Thus:

StringUtils.substring("abc", 0, 4) = "abc"

Solution 3:

Apache Commons Lang has a StringUtils.left method for this.

String upToNCharacters = StringUtils.left(s, n);