我想在许多字符之一（下面列出）上分割一个句子。我的正则表达式能够根据大多数字符进行拆分，但不能基于'['，']'（方括号的开头和结尾）进行拆分。如果我将字符串SPECIAL_CHARACTERS_REGEX更改为[ :;'=\\()!-\\[\\]]，它将开始在字符串中的整数上进行拆分，而不是在方括号中进行拆分。如何使正则表达式在方括号而不是整数（“ []”表示所有整数）上分割。
另一个相关的问题，是否有一种方法也可以从字符串中拆分数字？例如，9pm应分为9和pm。
This:
private static final String SPECIAL_CHARACTERS_REGEX = "[ :;'=\\()!-]";
String rawMessage = "let's meet tomorrow at 9:30p? 7-8pm? i=you go (no Go!) [to do !]"
String[] tokens = rawMessage.split(SPECIAL_CHARACTERS_REGEX);
Gives:
Input: let's meet tomorrow at 9:30p? 7-8pm? i=you go (no Go!) [to do !]
output: [let, s, meet, tomorrow, at, 9, 30p?, 7, 8pm?, i, you, go, , no, Go, , , [to, do, , ]]
和，
This:
private static final String SPECIAL_CHARACTERS_REGEX = "[ :;'=\\()!-\\[\\]]";
String rawMessage = "let's meet tomorrow at 9:30p? 7-8pm? i=you go (no Go!) [to do !]"
String[] tokens = rawMessage.split(SPECIAL_CHARACTERS_REGEX);
Gives:
let's meet tomorrow at 9:30p? 7-8pm? i=you go (no Go!) [to do !]
[let, s, meet, tomorrow, at, , , , , p, , , , , pm, , i, you, go, , no, , o, , , , to, do]
预期产量：
{"let", "s", "meet", "tomorrow", "at", "9", "30", "p", "7", "8", "pm", "i", "you", "go", "no", "Go", "to", "do"}
如果将破折号留在字符类的中间，则还需要对其进行转义。
但是，请避免将其放在角色类的开头或结尾。同样，您不需要在()这里转义，并且可能要在字符类之后*或+之后使用量词。
更新：为了获得预期的结果，您可以这样做。
private static final String SPECIAL_CHARACTERS_REGEX = "[ :;'?=()!\\[\\]-]+|(?<=\\d)(?=\\D)";
String rawMessage = "let's meet tomorrow at 9:30p? 7-8pm? i=you go (no Go!) [to do !]";
String[] tokens = rawMessage.split(SPECIAL_CHARACTERS_REGEX);
System.out.println(Arrays.toString(tokens));
正则表达式：
[ :;'?=()!\[\]-]+ any character of: ' ', ':', ';', ''', '?',
'=', '(', ')', '!', '\[', '\]', '-' (1 or more times)
| OR
(?<= look behind to see if there is:
\d digits (0-9)
) end of look-behind
(?= look ahead to see if there is:
\D non-digits (all but 0-9)
) end of look-ahead
输出
[let, s, meet, tomorrow, at, 9, 30, p, 7, 8, pm, i, you, go, no, Go, to, do]
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