java 从字符串中删除不在白名单中的所有字符
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Remove all characters from string which are not on whitelist
提问by PerwinCZ
I am trying to write java code which would remove all unwanted characters and let there be only whitelisted ones.
我正在尝试编写 Java 代码，该代码将删除所有不需要的字符，并只保留白名单中的字符。
Example:
例子：
String[] whitelist = {"a", "b", "c"..."z", "0"..."9", "[", "]",...}
I want there only letters (lower and uppercase) and numbers + some next characters I would add.
Then I would start for()cycle for every character in the string, and replace it with empty string if it isn't on whitelist.
我只想要字母（小写和大写）和数字 + 我要添加的一些下一个字符。然后我将对for()字符串中的每个字符开始循环，如果它不在白名单中，则用空字符串替换它。
But that isn't good solution. Maybe it could be done somehow using pattern (regex)? Thanks.
但这不是很好的解决方案。也许可以使用模式（正则表达式）以某种方式完成？谢谢。
回答by Jon Skeet
Yes, you can use String.replaceAllwhich takes a regex:
是的，您可以使用String.replaceAllwhich 需要正则表达式：
String input = "BAD good {} []";
String output = input.replaceAll("[^a-z0-9\[\]]", "");
System.out.println(output); // good[]
Or in Guavayou could use a CharMatcher:
或者在番石榴中，您可以使用CharMatcher：
CharMatcher matcher = CharMatcher.inRange('a', 'z')
.or(CharMatcher.inRange('0', '9'))
.or(CharMatcher.anyOf("[]"));
String input = "BAD good {} []";
String output = matcher.retainFrom(input);
That just shows the lower case version, making it easier to demonstrate. To include upper case letters, use "[^A-Za-z0-9\\[\\]]"in the regex (and any other symbols you want) - and for the CharMatcheryou can orit with CharMatcher.inRange('A', 'Z').
那只是显示小写版本，使其更易于演示。要包含大写字母，请"[^A-Za-z0-9\\[\\]]"在正则表达式（以及您想要的任何其他符号）中使用 - 对于CharMatcher您可以or使用CharMatcher.inRange('A', 'Z').
回答by Thomas
You could try and match everything that is not in your whitelist and replace it with an empty string:
您可以尝试匹配不在白名单中的所有内容并将其替换为空字符串：
String in = "asng $%& 123";
//this assumes your whitelist contains word characters and whitespaces, adapt as needed
System.out.println(in.replaceAll( "[^\w\s]+", "" ));
