正则表达式查找所有匹配项

Question

我需要一个正则表达式来查找我的模式的所有匹配项。

文本是这样的：

"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 [.*]

我已经这样尝试过：

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

与此匹配的文本是这样的：

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

但我期望如下：

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

请帮帮我，java中的正则表达式将如何获得上述输出。

Answer 1

@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]");
}

编辑

Trinimon回答的模式也许更优雅一些。我用他的正则表达式更新了正则表达式。

Answer 2

您需要使用非贪婪匹配：

start.*?end

在你的情况下，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());
}

证明链接：http : //ideone.com/Qtymb3