以星号文字结尾的单词的 Java 正则表达式模式匹配
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Java regex pattern matching for word ending with asterix literal
提问by Shamik
I'm trying to make a simple regex pattern work using java. I need to recognize any uppercase word ending with trailing asterisk with a sentence. From the following example :
我正在尝试使用 java 制作一个简单的正则表达式模式。我需要识别任何以尾随星号结尾的大写单词。从以下示例中：
Test ABC* array
String text = "Test ABC* array";
Matcher m = Pattern.compile("\b[A-Z]+[*]?\b").matcher(text);
String text = "Test ABC* array";
Matcher m = Pattern.compile("\b[A-Z]+[*]?\b").matcher(text);
Thanks
谢谢
回答by stema
the problem is that you don't have a word boundaryat the end after the star. So try this
问题是在星号之后的末尾没有单词边界。所以试试这个
Matcher m = Pattern.compile("\b[A-Z]+\*\B").matcher(text);
\Bis not a word boundary, so this is exactly what you get between the *and a whitespace.
\B不是单词边界，因此这正是您在*空格和空格之间得到的。
See it here on Regexr
在 Regexr 上看到它
回答by flawyte
Assuming your string can contain multiple AAA* parts :
假设您的字符串可以包含多个 AAA* 部分：
String text = "Test ABC* array";
Matcher m = Pattern.compile("([A-Z]+\*)").matcher(text);
while (m.find()) {
System.out.println(m.group());
}
回答by David Sonnenfeld
I would keep it simple:
我会保持简单：
if(text.contains("*")) {
int index = text.lastIndexOf("*");
String ident= text.substring(0,index-1);
}
回答by jimbo
You have to escape the literal asterisk with a backslash. It ends up being a double-backslash in Java.
您必须使用反斜杠来转义字面的星号。它最终成为 Java 中的双反斜杠。
"[A-Z]+\*"
