问题描述
我发现了一些关于正则表达式过滤掉非英语的引用，但是Java中的，除了它们都指的是问题比我想要解决的问题：
I found a few references to regex filtering out non-English but none of them is in Java, aside from the fact that they are all referring to somewhat different problems than what I am trying to solve:
- 替换所有非英文字符带有空格的
。 - 创建一个返回
true的方法
如果字符串包含任何非英语
字符。
英文文本不仅指实际字母和数字，还指标点符号。
By "English text" I mean not only actual letters and numbers but also punctuation.
到目前为止，我能够为目标＃1带来的非常简单：
So far, what I have been able to come with for goal #1 is quite simple:
String.replaceAll("\\W", " ")
实际上，这么简单，我怀疑我错过了什么......您是否发现上述任何警告？
In fact, so simple that I suspect that I am missing something... Do you spot any caveats in the above?
至于目标＃2，我可以简单地 trim（） 上面的 replaceAll（）之后的字符串，然后检查它是否为空。但是......有更有效的方法吗？
As for goal #2, I could simply trim() the string after the above replaceAll(), then check if it's empty. But... Is there a more efficient way to do this?
推荐答案
\ W 相当于 [^ \w] 和 \w 相当于 [a-zA-Z_0-9] 。使用 \W 将替换所有，这不是字母，数字或下划线—喜欢标签和换行符。这个问题是否真的取决于你。
\W is equivalent to [^\w], and \w is equivalent to [a-zA-Z_0-9]. Using \W will replace everything which isn't a letter, a number, or an underscore — like tabs and newline characters. Whether or not that's a problem is really up to you.
在这种情况下，您可能希望使用省略标点符号的字符类;类似
In that case, you might want to use a character class which omits punctuation; something like
[^\w.,;:'"]
使用和。
Pattern p = Pattern.compile("\\W");
boolean containsSpecialChars(String string)
{
Matcher m = p.matcher(string);
return m.find();
}
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