Java如何用字符串中的单个空格替换2个或多个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、简单的方法来改变这个字符串

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，除了我还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.

这样的事情让我部分地在那里

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(" +", " ");

另见

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

  也可以只用一个 replaceAll 来做到这一点，但这比 trim() 解决方案可读性要差得多.尽管如此，这里提供它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 个备选:

  • ^_+ : 字符串开头的任意空格序列
    • 匹配并替换为$1，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
      • 匹配并替换为$1，捕获空字符串
      • Match and replace with $1, which captures the empty string
      • 匹配并替换为 $1，捕获一个空格
      • Match and replace with $1, which captures a single space
      • regular-expressions.info/Anchors

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