What's the simplest way to get string between specific two characters?

Question

For example, I have front boundary 'x' and back boundary 'y'

EG Given a string 'abcxfoobaryblablah', it should return 'foobar'.

Assume the two boundaries only appear exactly once in the string like in the example.

Thanks! Doesn't have to use Regex, but I guess that's the best method.

Answer 1

String s = "abcxfoobaryblablah";
s = s.substring(s.indexOf('x') + 1, s.indexOf('y'));
System.out.println(s);

Out

foobar

Answer 2

查看Java的边界匹配器http://docs.oracle.com/javase/tutorial/essential/regex/bounds.html

Answer 3

May be a long code but just a simple logic.

String str="abcxfoobaryblablah";
char array[]=str.toCharArray();
boolean flag=false;
String answer="";

for(int i=0;i<array.length;i++)
{
if(array[i]=='x'){
    flag=true;
    continue;
}
if(array[i]=='y') break;
if(flag)answer+=array[i];
}

System.out.println(answer);

Answer 4

String  input = "abcxfoobaryblablah";

int startIndex = input.indexOf("x");

int endIndex =  input.indexOf("y");     



String output = input.substring(startIndex+1, endIndex);

System.out.println(output);

Answer 5

You could certainly do substrings, but it is possible with regex.

String str = "abcxfoobaryblablah";
str = str.replaceAll("^.*x|y.*$", "");

^.*x matches the characters from the start of the String up to x . y.*$ matches the characters from y to the end of the String .

Note that if x is after y , then this will keep just the characters after x to the end of the String . If x is not present, it will keep the characters from the start of the String to y . If you only want it to work when both x and y are in there in the correct order, then this is what you want:

String str = "abcxfoobaryblablah";
str = str.replaceAll("^.*x(.*)y.*$", "$1");

This uses groups .

Answer 6

最简单的方法是：

String middle = str.replaceAll(".*x(.*)y.*", "$1");