Question

我的模式如下

API:ADD|TYPE:ABC...MATCH:TRUE

[LOTS OF OTHER LOG LINES]

API:ADD|TYPE:ABC...MATCH:TRUE

[LOTS OF OTHER LOG LINES]

API:ADD|TYPE:DEF...MATCH:TRUE

我尝试了以下正则表达式：

(API:.*MATCH:(TRUE|FALSE))

while (matcher.find()) {
    System.out.println(i + " occurence");
    i++;
    matches.add(matcher.group());
}

它从第一个“API”匹配到最后一个“TRUE”，因此只返回一个子字符串！我想要三个子串（在这种情况下）从“API”开始直到“TRUE”或“FALSE”。

感谢你在这方面的帮助。谢谢。

修改

-------------------------------------------------------
20:31:57    CALL    add     35  
-------------------------------------------------------
20:31:57    REASON  API:ADD|TYPE:ABC|ErrorType:VALIDATION|Error Message:User already has|MATCH:FALSE

Answer 1

使用非贪婪量词.*?，正则表达式将尽可能少地开始匹配。

(API:.*?MATCH:(TRUE|FALSE))

Answer 2

This is because you use .* which is a greedy quantifier.

You should try to use .*? which is reluctant -- that means that it matches the smallest possible substring.

For more info about greedy vs. reluctant see the excellent answers here: Greedy vs. Reluctant vs. Possessive Quantifiers

Java - 给定正则表达式的所有子字符串

2 个答案: