正则表达式通过用运算符（！，＆，（（），|）拆分所有字符串的开始和结束位置来获取String数组在Java中

Question

I am new to regex , I want to get string array with all string's start position and end position by splitting operator ( ! , & , ( , ) , | ) .

Examples:

1.  !(AString&BString)|CString
    0123456789

Output should be String Array with all string's start position and end position: 
[AString,BString,CString...]
Every String Position:
AString (7 length) => 2 to 8
BString => 10 to 16
CString => 19 to 25


2.  (AString)&(BString)|!(CString)
3.  !(AString|BString)&CString
4.  !(AString&(!(BString|CString))|DString)

Answer 1

Instead of splitting, you could do matching. [^()!&|]+ matches any character but not of ( or ) or | or ! or & one or more times. Then find the start and end index of each match using matcherobj.start() and matcherobj.end() functions.

String s1 = "!(AString&BString)|CString";
Matcher m = Pattern.compile("[^()!&|]+").matcher(s1);
while(m.find())
{

System.out.println(m.group() + "  => " + m.start() +  " to " + (m.end()-1));

}

Output:

AString  => 2 to 8
BString  => 10 to 16
CString  => 19 to 25

Answer 2

You can use Strings Split method:

String givenString= "!(AString&BString)|CString";
String[] stringArray= givenString.split("[!()&|]");//modify regular expression according to your need 

This will give me your array which contains element as you wanted plus other element as "" .