Question

echo [18%] | sed s:[\[%\]]::g

我真的对此感到困惑，因为同样的模式成功地取代了vim中的[18%]。我还在一些在线正则表达式工具中对表达式进行了测试，他们都表示它会按照预期在[，%和]上匹配。我尝试添加-r选项以及在引号中包含替换命令。

我知道我可以使用其他命令来完成这项任务，但我想知道它为什么会这样做，这样我才能更好地理解sed。

Answer 1

$ echo [18%] | sed s:[][%]::g
18

sed支持POSIX.2正则表达式语法：默认情况下为基本（BRE）语法，带有-r标志的扩展语法。在POSIX.2语法（基本或扩展）中，通过使其成为字符类中的第一个字符，可以包含右方括号。反斜杠无济于事。

这很烦人，因为几乎所有其他现代语言和工具都使用Perl或类似Perl的正则表达式语法。 POSIX语法是不合时宜的。

您可以在regex（7）手册页中阅读有关POSIX.2语法的内容。

 A bracket expression is a list of  characters  enclosed  in  "[]".   It  normally
 matches  any  single character from the list (but see below).  If the list begins
 with '^', it matches any single character (but see below) not from  the  rest  of
 the  list.  If two characters in the list are separated by '-', this is shorthand
 for the full range of characters between those two (inclusive) in  the  collating
 sequence,  for  example, "[0-9]" in ASCII matches any decimal digit.  It is ille‐
 gal(!) for two ranges to share an endpoint, for  example,  "a-c-e".   Ranges  are
 very  collating-sequence-dependent, and portable programs should avoid relying on
 them.

 To include a literal ']' in the list, make it the first  character  (following  a
 possible '^').  To include a literal '-', make it the first or last character, or
 the second endpoint of a range.  To use a literal '-' as the first endpoint of  a
 range,  enclose  it in "[." and ".]"  to make it a collating element (see below).
 With the exception of these and some  combinations  using  '['  (see  next  para‐
 graphs), all other special characters, including '\', lose their special signifi‐
 cance within a bracket expression.

为什么这个sed命令输出“[18”而不是“18”？

1 个答案: