Java Regex在分隔符和保持分隔符之间拆分字符串

Question

我有一个看起来像这样的java字符串;

(fname:jon)(lname:doe)(guaranteer: Sam (W) Willis)(age:35)(addr:1 Turnpike Plaza)(favcolor:blue)

我想从分隔符中拆分这个String (fname:jon)<here>(lname:doe).

我尝试通过正则表达式进行拆分,\)\(但它只是破坏了我的代码

arr = s.split("\\)\\(");
for (String a: arr) System.out.println(a);

产量

(fname:jon
lname:doe
guaranteer: Sam (W) Willis
age:35
addr:1 Turnpike Plaza
favcolor:blue)

我也看了这个问题:如何拆分字符串,还要保留分隔符？,但它没有帮助,因为在我的情况下,我想要)(均匀地保持分隔符和分割分隔符,即,第一个括号应该是第一个结果,第二个到第二个结果.

我使用的正则表达式是s.split("(?<=\\)\\()")它给出了输出:

(fname:jon)(
lname:doe)(
guaranteer: Sam (W) Willis)(
age:35)(
addr:1 Turnpike Plaza)(
favcolor:blue)

这是我想要的输出:

(fname:jon)
(lname:doe)
(guaranteer: Sam (W) Willis)
(age:35)
(addr:1 Turnpike Plaza)
(favcolor:blue)

Answer 1

您可以使用正向前瞻和正向后视分割您的字符串,如下所示:

正则表达式 (?<=\))(?=\()

DEMO

1. (?<=\)) positive lookbehind表示结束括号应位于分割位置之前.
2. (?=\() 积极前瞻,表明开口括号应该遵循该分割位置.

产量

(fname:jon)
(lname:doe)
(guaranteer: Sam (W) Willis)
(age:35)
(addr:1 Turnpike Plaza)
(favcolor:blue)

码

String s = "(fname:jon)(lname:doe)(guaranteer: Sam (W) Willis)(age:35)(addr:1 Turnpike Plaza)(favcolor:blue)";
String arr[] = s.split("(?<=\\))(?=\\()");
for (String a: arr) System.out.println(a);

Answer 2

你可以使用积极的lookbehind和负向前瞻来实现这个目标:

public static void main(String[] args) throws Exception {
    String s = "(fname:jon)(lname:doe)(guaranteer: Sam (W) Willis)(age:35)"
            + "(addr:1 Turnpike Plaza)(favcolor:blue)";
    String[] arr = s.split("(?<=\\)(?!\\s))"); // positive lookbehind for ")" and negative lookahead for "space"
    for (String str : arr) {
        System.out.println(str);
    }

}

O/P:

(fname:jon)
(lname:doe)
(guaranteer: Sam (W) Willis)
(age:35)
(addr:1 Turnpike Plaza)
(favcolor:blue)