正则表达式找到所有匹配

我需要一个正则表达式来查找我的模式的所有匹配项 .

文字是这样的:

"someother text !style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9] end of line text"

我想找到该模式的所有匹配项:

!style_delete [.*]

我试过这样的:

Pattern pattern = Pattern.compile("!style_delete\\s*\\[.*\\]");

有了这个匹配文本就像这样:

!style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9]

但我预计如下:

match 1 : !style_delete [company code : 43ev4] 
match 2 : !style_delete [organiztion : 0asj9]

请帮助我,java中的正则表达式将获得以上输出 .

回答(3)

2 years ago

您需要使用non-greedy匹配:

start.*?end

在您的情况下,模式是:

!style_delete\\s\\[(.*?)\\] (Even simple to understand than first version :))

证明(Java 7):

String string = "someother text !style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9] end of line text"; 
Pattern pattern = Pattern.compile("!style_delete\\s\\[(.*?)\\]");
Matcher matcher = pattern.matcher(string) ;
while (matcher.find()) {
    System.out.println(matcher.group());
}

链接到证据:http://ideone.com/Qtymb3

2 years ago

@Test
public void test() {
    final String input = "someother text !style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9] end of line text";
    // my regexp:strong text
    // final String regex = "(!style_delete\\s\\[[a-zA-Z0-9\\s:]*\\])";
    // regexp from Trinmon:
    final String regex = "(!style_delete\\s*\\[[^\\]]*\\])";

    final Matcher m = Pattern.compile(regex).matcher(input);

    final List<String> matches = new ArrayList<>();
    while (m.find()) {
        matches.add(m.group(0));
    }

    assertEquals(2, matches.size());
    assertEquals("match 1: ", matches.get(0), "!style_delete [company code : 43ev4]");
    assertEquals("match 2: ", matches.get(1), "!style_delete [organiztion : 0asj9]");
}

edit

也许Trinimon的答案模式更优雅一点 . 我用正则表达式更新了正则表达式 .

2 years ago

这是因为 .* 贪婪 . 改为使用它:

"!style_delete\\s*\\[[^\\]]*\\]"

这意味着:匹配括号中的所有内容,不包括结束 ] .

或者使 [] 之间的内容非贪婪:

"!style_delete\\s*\\[.*?\\]"