Java如何用字符串中的单个空格替换2个或更多空格并删除前导和尾随空格 [英] Java how to replace 2 or more spaces with single space in string and delete leading and trailing spaces

问题描述

在Java中寻找快速，简单的方法来更改此字符串

Looking for quick, simple way in Java to change this string

" hello     there   "

看起来像这样的东西

"hello there"

我用一个空格替换所有这些多个空格，除了我也希望字符串开头的一个或多个空格消失。

where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone.

这样的东西让我部分地在那里

Something like this gets me partly there

String mytext = " hello     there   ";
mytext = mytext.replaceAll("( )+", " ");

但不完全。

推荐答案

试试这个：

String after = before.trim().replaceAll(" +", " ");

参见

• String.trim（）
  
  
  
  
  • 返回字符串的副本，省略前导和尾随空格。
  
  

  只需一个 replaceAll ，但这比 trim（）解决方案的可读性低得多。尽管如此，这里提供的只是为了展示正则表达式可以做什么：

  It's also possible to do this with just one replaceAll, but this is much less readable than the trim() solution. Nonetheless, it's provided here just to show what regex can do:

      String[] tests = {
          "  x  ",          // [x]
          "  1   2   3  ",  // [1 2 3]
          "",               // []
          "   ",            // []
      };
      for (String test : tests) {
          System.out.format("[%s]%n",
              test.replaceAll("^ +| +$|( )+", "$1")
          );
      }
  

  有3个替代品：

  
  
  • ^ _ + ：字符串开头的任何空格序列
    
    
    
    
    • 匹配并替换为 $ 1 ，它捕获空字符串
    
    
    • ^_+ : any sequence of spaces at the beginning of the string
      • Match and replace with $1, which captures the empty string
      
      
      • 匹配和替换为 $ 1 ，它捕获空字符串
      
      
      • Match and replace with $1, which captures the empty string
      
      
      • 匹配和替换使用 $ 1 ，捕获单个空格
      
      
      • Match and replace with $1, which captures a single space
      
      
      • regular-expressions.info/Anchors
      
      
      • regular-expressions.info/Anchors

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