从Java字符串中去除前导和尾随空格

Question

是否有一种方便的方法可以从Java字符串中剥离任何前导或尾随空格？

就像是：

String myString = "  keep this  ";

String stripppedString = myString.strip();

System.out.println("no spaces:" + strippedString);

结果：

no spaces:keep this

myString.replace(" ","") 将替换keep和this之间的空间。

Answer 1

您可以尝试trim（）方法。String newString = oldString.trim();看看javadocs

Answer 2

trim（）是您的选择，但是如果您想使用replacemethod －可能更灵活，则可以尝试以下操作：String stripppedString = myString.replaceAll("(^ )|( $)", "");

Answer 3

现在，使用java-11，您可以利用String.stripAPI返回一个值为该字符串的字符串，并删除所有前导和尾随空格。相同的javadoc读取：/**&nbsp;* Returns a string whose value is this string, with all leading&nbsp;* and trailing {@link Character#isWhitespace(int) white space}&nbsp;* removed.&nbsp;* <p>&nbsp;* If this {@code String} object represents an empty string,&nbsp;* or if all code points in this string are&nbsp;* {@link Character#isWhitespace(int) white space}, then an empty string&nbsp;* is returned.&nbsp;* <p>&nbsp;* Otherwise, returns a substring of this string beginning with the first&nbsp;* code point that is not a {@link Character#isWhitespace(int) white space}&nbsp;* up to and including the last code point that is not a&nbsp;* {@link Character#isWhitespace(int) white space}.&nbsp;* <p>&nbsp;* This method may be used to strip&nbsp;* {@link Character#isWhitespace(int) white space} from&nbsp;* the beginning and end of a string.&nbsp;*&nbsp;* @return&nbsp; a string whose value is this string, with all leading&nbsp;*&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; and trailing white space removed&nbsp;*&nbsp;* @see Character#isWhitespace(int)&nbsp;*&nbsp;* @since 11&nbsp;*/public String strip()这些示例案例可能是：-System.out.println("&nbsp; leading".strip()); // prints "leading"System.out.println("trailing&nbsp; ".strip()); // prints "trailing"System.out.println("&nbsp; keep this&nbsp; ".strip()); // prints "keep this"