如何删除一个字符串在Java中，只有尾部的空格，并保持领先的空间？ [英] How to remove only trailing spaces of a string in Java and keep leading spaces?

问题描述

的<一个href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#trim%28%29"><$c$c>trim()功能将同时删除尾随和前导空格，但是，如果我只是想删除一个字符串的尾部空间，我该怎么办呢？

The trim() function removes both the trailing and leading space, however, if I only want to remove the trailing space of a string, how can I do it?

谢谢！

推荐答案

使用常规的前pression \ S + $ ，可以更换所有尾随空格字符（包括空格和制表符）没事（，）。

Using the regular expression \s+$, you can replace all trailing space characters (includes space and tab characters) with nothing ("").

final String text = "  foo   ";
System.out.println(text.replaceFirst("\\s+$", ""));

下面是正则表达式的分解：

Here's a breakdown of the regex:

• \ S ＆ndash的;任何空白字符
• + ＆ndash的;匹配一个或多个所述previous令牌;即，匹配一个或多个空格字符
• $ ＆ndash的;所述字符串的末尾

• \s – any whitespace character
• + – match one or more of the previous token; i.e., match one or more whitespace character
• $ – the end of the string

因此​​，定期前pression将匹配尽可能多的空白，因为它可以是直接后跟字符串的末尾。换言之，后空白

Thus, the regular expression will match as much whitespace as it can that is followed directly by the end of the string: in other words, the trailing whitespace.

该投资进入学习正规的前pressions将变得更有价值，如果你需要扩展后您的要求。

The investment into learning regular expressions will become more valuable, if you need to extend your requirements later on.

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