如何在正则表达式中找到行尾？

Question

I have the following regex in = in.replaceAll(" d+\\n", "");

I wanted to use it to get rid of the "d" at the end of lines

But I just won't do that d
<i>I just won't do that</i> d

No, no-no-no, no, no d

What is not accurate with my regex in = in.replaceAll(" d+\\n", "");

Answer 1

Most probably your lines are not separated only with \\n but with \\r\\n . You can try with \\r?\\n to optionally add \\r before \\n . Lets also not forget about last b which doesn't have any line separators after it. To handle it you need to add $ in your regex which means anchor representing end of your data. So your final pattern could look like

in.replaceAll(" d+(\r?\n|$)", "")

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In case you don't want to remove these line separators you can use "end of line anchor" $ with MULTILINE flag (?m) instead of line separators like

in.replaceAll("(?m) d+$", "")

especially because there are no line separators after last b .

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In Java, when MULTILINE flag is specified, $ will match the empty string:

• Before a line terminator :
  • A carriage-return character followed immediately by a newline character ( "\\r\\n" )
  • Newline (line feed) character ( '\\n' ) without carriage-return ( '\\r' ) right in front
  • Standalone carriage-return character ( '\\r' )
  • Next-line character ( '\…' )
  • Line-separator character ( '\
' )
  • Paragraph-separator character ( '\
' )
• At the end of the string

When UNIX_LINES flag is specified along with MULTILINE flag, $ will match the empty string right before a newline ( '\\n' ) or at the end of the string.

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Anyway if it is possible don't use regex with HTML .

Answer 2

As Pshemo states in his answer , your string most likely contains Windows-style newline characters, which are \\r\\n as opposed to just \\n .

You can modify your regex to account for both newline character (plus the case where the string ends with a d without a newline) with the code:

in = in.replaceAll("(d+(?=\r\n)|d+(?=\n)|d+$)","");

This regex will remove anything that matches d+ followed by \\r\\n , d+ followed by \\n or d+$ (any d before the end of the String).

(d+(?=\\r\\n)|d+(?=\\n)|d+$)

Debuggex Demo