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从给定字符串替换\\ n \\ r和\\ t的最佳方法

[英]Optimal way to replace \n \r and \t from a given String

Is there a better way to do this in Java: 在Java中是否有更好的方法可以做到这一点:

public static String clearLog(String message) {
    return message.replace('\n', ' ').replace('\r', ' ').replace( '\t' , ' ');
}

I know this creates 3 String object and I want to avoid it. 我知道这会创建3 String对象,我想避免它。

You can use String::replaceAll like this : 您可以像这样使用String::replaceAll

message.replaceAll("[\n\r\t]", " ");

because replaceAll uses regex, so you can create a class which holds these three characters [\\n\\r\\t] replacing \\n or \\r or \\t with a space. 因为replaceAll使用正则表达式,所以您可以创建一个包含这三个字符[\\n\\r\\t]的类,并用空格替换\\n\\r\\t

You can do it without regex using a char[] : 您可以使用char[]而不使用正则表达式:

char[] cs = message.toCharArray();
for (int a = 0; a < cs.length; ++a) {
  switch (cs[a]) {
    case '\n': case '\r': case '\t':
      cs[a] = ' ';
      break;
  }
}
return new String(cs);

This will likely be a lot faster than the regex approach because it's very simple code to execute: it doesn't involve the whole regex engine; 这可能比regex方法快很多,因为它是非常简单的代码执行:它不涉及整个regex引擎。 but it's more verbose, less readable and less flexible. 但是它比较冗长,可读性和灵活性较差。

Sure, use a Guava CharMatcher . 当然,请使用Guava CharMatcher

String cleaned = CharMatcher.anyOf("\t\n\r").trimAndCollapseFrom(yourString, ' ');

CharMatcher is heavily optimized and will limit object creation. CharMatcher经过了严格的优化,将限制对象的创建。 If you save the CharMatcher to a constant (you can, it's immutable), then this will only generate one interim Object, a StringBuilder . 如果将CharMatcher保存为一个常量(可以,它是不​​可变的),那么它将仅生成一个临时对象StringBuilder

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