java Java中只包含字母、数字和空格的字符串的正则表达式
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Regexp for a string to contain only letters , numbers and space in Java
提问by sat
Requirement: String should contain only letters , numbers and space.
I have to pass a clean name to another API.
要求：字符串只能包含字母、数字和空格。
我必须将一个干净的名称传递给另一个 API。
Implementation: Java
实现：Java
I came up with this for my requirement
我想出了这个以满足我的要求
public static String getCleanFilename(String filename) {
if (filename == null) {
return null;
}
return filename.replaceAll("[^A-Za-z0-9 ]","");
}
This works well for few of my testcase , but want to know am I missing any boundary conditions, or any better way (in performance) to do it.
这适用于我的少数测试用例，但想知道我是否缺少任何边界条件，或任何更好的方法（在性能方面）来做到这一点。
采纳答案by Stefan Kendall
To answer you're direct question, \tfails your method and passes through as "space." Switch to \s([...\s]and you're good.
要回答您的直接问题，\t请使您的方法失败并作为“空间”通过。切换到\s([...\s]你很好。
At any rate, your design is probably flawed. Instead of arbitrarily dicking with user input, let the user know what you don't allow and make the correction manual.
无论如何，您的设计可能存在缺陷。与其随意修改用户输入，不如让用户知道您不允许什么并制作更正手册。
EDIT:
If the filename doesn't matter, take the SHA-2 hash of the file name and use that. Guaranteed to meet your requirements.
编辑：
如果文件名无关紧要，请使用文件名的 SHA-2 哈希值并使用它。保证满足您的要求。
回答by Howard
Additional to comments: i don't think that performance is an issue in a scenario where user input is taken (and a filename shouldn't be that long...).
补充评论：我认为在采用用户输入的情况下性能不是问题（并且文件名不应该那么长......）。
But concerning your question: you may reduce the number of replacements by adding an additional + in your regex:
但是关于您的问题：您可以通过在正则表达式中添加额外的 + 来减少替换次数：
[^A-Za-z0-9 ]+
[^A-Za-z0-9 ]+
