如何过滤Java String以仅获取字母字符？ [英] How to filter a Java String to get only alphabet characters?

问题描述

我正在生成一个XML文件来进行付款，我对用户的全名有一个约束。那个参数只接受字母字符（a-ZAZ）+空格来分隔名字和姓氏。

I'm generating a XML file to make payments and I have a constraint for user's full names. That param only accept alphabet characters (a-ZAZ) + whitespaces to separe names and surnames.

我无法以简单的方式过滤这个，我怎么能构建正则表达式或过滤器以获得我想要的输出？

I'm not able to filter this in a easy way, how can I build a regular expression or a filter to get my desireable output?

示例：

'CarmenLópez-Delina Santos'必须'Carmen LopezDelina Santos'

我需要用单个元音转换带有装饰的元音，如下所示：á> a，à> a，a，等等;并删除点，连字符等特殊字符。

I need to transform vowels with decorations in single vowels as follows: á > a, à > a, â > a, and so on; and also remove special characters as dots, hyphens, etc.

谢谢！

推荐答案

您可以先使用规范化程序，然后删除不受欢迎的字符：

You can first use a Normalizer and then remove the undesired characters:

String input = "Carmen López-Delina Santos";
String withoutAccent = Normalizer.normalize(input, Normalizer.Form.NFD);
String output = withoutAccent.replaceAll("[^a-zA-Z ]", "");
System.out.println(output); //prints Carmen LopezDelina Santos

请注意，这可能不适用于所有和任何非ascii字母任何语言 - 如果遇到这种情况，该信件将被删除。一个这样的例子是土耳其语 i 。

Note that this may not work for all and any non-ascii letters in any language - if such a case is encountered the letter would be deleted. One such example is the Turkish i.

在这种情况下的替代方案可能是列出所有可能的字母和他们的替代......

The alternative in that situation is probably to list all the possible letters and their replacement...

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