java 正则表达式忽略前导字符
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Regex to ignore leading characters
提问by wobblycogs
I'm sure there's a simple solution to this but it's got me stuck so hopefully someone can give me a hand. I've got multiple lines of input that look like this:
我确信有一个简单的解决方案，但它让我陷入困境，所以希望有人能帮我一把。我有多行输入，如下所示：
FooABC
Foo XYZ123
FooFgh
I need a regex (in Java) that will give me:
我需要一个正则表达式（在 Java 中），它会给我：
ABC
XYZ123
Fgh
The best I've come up with so far is:
到目前为止我想出的最好的是：
[^Foo\s?].*$
Which I thought was working until I noticed that it was removing the capital F from Fgh. Is there any way I can make it exactly ignore the first three characters and one or more whitespace characters then just capture everything else?
我认为这是有效的，直到我注意到它正在从 Fgh 中删除大写 F。有什么办法可以让它完全忽略前三个字符和一个或多个空白字符，然后只捕获其他所有字符？
EDIT / CLARIFICATION
编辑/澄清
I should have said that I can't use substring / replace or any of the things I would normally do in this situation because this regex will be fed into deployed code that can't currently be updated. What the code is doing is this:
我应该说我不能使用子字符串/替换或在这种情况下我通常会做的任何事情，因为这个正则表达式将被输入到当前无法更新的部署代码中。代码在做什么是这样的：
Pattern pattern = Pattern.compile( regex );
Matcher matcher = pattern.matcher( data );
matcher.find();
String extracted = matcher.group();
and it's then using the extracted string as a key for a look up. So it has to extract the key in one pass.
然后它使用提取的字符串作为查找的键。所以它必须一次性提取密钥。
回答by kennytm
Replace
代替
^Foo\s*
with an empty string. Use .{3}if you need 3 arbitrary characters instead of Foo.
带有空字符串。使用.{3}，如果你需要3个任意字符来代替Foo。
If you need to use match instead of replace, you need to use a capture group:
如果需要使用匹配而不是替换，则需要使用捕获组：
^Foo\s*(.*)$
The [^Foo\s?]is a character class ([...]) to match any characters except (^)an F, a o, a whitespace (\s) or a question mark ?. That's why the Fin Fghis also matched. If you want a grouping bracket, use (?:...).
的[^Foo\s?]是一个字符类（[...]）来匹配任何字符除了（^）的F，一o，一个空格（\s）或问号?。这就是FinFgh也匹配的原因。如果需要分组括号，请使用(?:...).
回答by Frank Schmitt
^.{3}\s+.*$
should do the trick (it matches 3 arbitrary characters at the beginning of the line, one or more whitespaces and then anything up until the end of the line).
应该可以解决问题（它匹配行首的 3 个任意字符，一个或多个空格，然后匹配行尾的任何字符）。
回答by Peter Lawrey
What about
关于什么
String text = text.substring(3).trim();
or KennyTM's solution
或 KennyTM 的解决方案
String text = text.replaceFirst("^Foo\s*", "");
