java java正则表达式过滤掉非英文文本
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java regex to filter out non-English text
提问by Regex Rookie
I found a few references to regex filtering out non-English but noneof them is in Java, aside from the fact that they are all referring to somewhat differentproblems than what I am trying to solve:
我发现了一些对 regex 过滤掉非英语的引用，但它们都不是在 Java 中的，除了它们都指的是与我试图解决的问题有些不同的问题：
- Replace all non-English characters with a space.
- Create a method that returns
trueif a string contains any non-English character.
- 用空格替换所有非英文字符。
- 创建一个方法，该方法
true在字符串包含任何非英语字符时返回。
By "English text" I mean not only actual letters and numbers but also punctuation.
“英文文本”不仅指实际的字母和数字，还指标点符号。
So far, what I have been able to come with for goal #1 is quite simple:
到目前为止，我能够为目标 #1 带来的东西非常简单：
String.replaceAll("\W", " ")
In fact, so simple that I suspect that I am missing something... Do you spot any caveats in the above?
事实上，如此简单以至于我怀疑我遗漏了什么......你在上面发现任何警告吗？
As for goal #2, I could simply trim()the string afterthe above replaceAll(), then check if it's empty. But... Is there a more efficient way to do this?
至于目标＃2，我可以简单trim()的字符串后上面replaceAll()，然后检查它是否是空的。但是......有没有更有效的方法来做到这一点？
回答by Matt Ball
In fact, so simple that I suspect that I am missing something... Do you spot any caveats in the above?
事实上，如此简单以至于我怀疑我遗漏了什么......你在上面发现任何警告吗？
\Wis equivalent to [^\w], and \wis equivalent to [a-zA-Z_0-9]. Using \Wwill replace everythingwhich 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.
\W等价于[^\w]，\w等价于[a-zA-Z_0-9]。Using\W将替换所有不是字母、数字或下划线的东西——比如制表符和换行符。这是否是一个问题真的取决于你。
By "English text" I mean not only actual letters and numbers but also punctuation.
“英文文本”不仅指实际的字母和数字，还指标点符号。
In that case, you might want to use a character class which omits punctuation; something like
在这种情况下，您可能希望使用省略标点符号的字符类；就像是
[^\w.,;:'"]
Create a method that returns true if a string contains any non-English character.
如果字符串包含任何非英语字符，则创建一个返回 true 的方法。
Pattern p = Pattern.compile("\W");
boolean containsSpecialChars(String string)
{
Matcher m = p.matcher(string);
return m.find();
}
回答by Eli Mashiah
Here is my solution. I assume the text may contain English words, punctuation marks and standard ascii symbols such as #, %, @ etc.
这是我的解决方案。我假设文本可能包含英文单词、标点符号和标准的 ascii 符号，例如 #、%、@ 等。
private static final String IS_ENGLISH_REGEX = "^[ \w \d \s \. \& \+ \- \, \! \@ \# \$ \% \^ \* \( \) \; \\ \/ \| \< \> \\" \' \? \= \: \[ \] ]*$";
private static boolean isEnglish(String text) {
if (text == null) {
return false;
}
return text.matches(IS_ENGLISH_REGEX);
}回答by Gil SH
This works for me
这对我有用
private static boolean isEnglish(String text) {
CharsetEncoder asciiEncoder = Charset.forName("US-ASCII").newEncoder();
CharsetEncoder isoEncoder = Charset.forName("ISO-8859-1").newEncoder();
return asciiEncoder.canEncode(text) || isoEncoder.canEncode(text);
}
回答by dogbane
Assuming an english word is made up of characters from: [a-zA-Z_0-9]
假设一个英文单词由以下字符组成：[a-zA-Z_0-9]
To return true if a string contains any non-English character, use string.matches:
要在字符串包含任何非英语字符时返回 true，请使用string.matches：
return !string.matches("^\w+$");
