英文:
Java RegEx split after every third whitespace
问题
Let's say I have the String "a 1 b c 2 3 d e f 4 5 6" and I would like to split it on every third white space to get the substrings "a 1 b", "c 2 3", "d e f", and so on. How would I do that using Regular Expressions? I have tried different variants of "(?=[\\w+][\\s+][\\w+][\\s+][\\w+][\\s+])" none of which seem to be right. What would be the correct RegEx?
英文:
Let´s say I have the String "a 1 b c 2 3 d e f 4 5 6" and I would like to split it on every third whitespace to get the substrings "a 1 b", " c 2 3", " d e f", and so on. How would I do that usuing Regular Exprssions? I have tried different variants of "(?=[\\w+][\\s+][\\w+][\\s+][\\w+][\\s+])" none of which seem to be right. What would be the correct RegEx?
答案1
得分: 3
使用正则表达式模式的模式匹配器：
\S+(?:\s+\S+){0,2}
这将根据您的定义匹配输入中的每个术语：
\S+ 匹配第一个“单词”
(?:
\s+ 空白字符
\S+ 另一个单词
){0,2} 重复零到两次
Java 代码示例：
String input = "a 1 b c 2 3 d e f 4 5 ";
String pattern = "\\S+(?:\\s+\\S+){0,2}";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(input);
List<String> matches = new ArrayList<>();
while (m.find()) {
matches.add(m.group(0));
}
System.out.println(matches);
这将打印：
[a 1 b, c 2 3, d e f, 4 5]
英文:
Use a pattern matcher with the regex pattern:
<!-- language: regex -->
\S+(?:\s+\S+){0,2}
This will match each term in the input according to your definition:
<!-- language: regex -->
\S+ match first "word"
(?:
\s+ whitespace
\S+ another word
){0,2} zero to two times
<!-- language: java -->
String input = "a 1 b c 2 3 d e f 4 5 ";
String pattern = "\\S+(?:\\s+\\S+){0,2}";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(input);
List<String> matches = new ArrayList<>();
while (m.find()) {
matches.add(m.group(0));
}
System.out.println(matches);
This prints:
<!-- language: none -->
[a 1 b, c 2 3, d e f, 4 5]
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