bash - 如何定义单词的参数？

Question

我想做的是在每个字符前后获取 2 个字符串。

输入文件:

hello

reader

.....

预期的输出是:

# # h e l //before character h is null and assign with '#". After character h are "e" and "l".
# h e l l //before character e is "h". After character e are "l" and "l".
h e l l o //before character l are "h" and "e". After character l are "l" and "o".
e l l o # //before character l are "e" and "l". After character l is "o".
l l o # # //before character o are "l" and "l". After character o is null and assign with '#".

# # r e a
# r e a d
r e a d e
e a d e r
a d e r #
d e r # #

这是代码:归功于 RudiC

awk '
        { L = length * 2
          M = int (L / 4)
          X = sprintf ("%*sY%*s", M, "", M, "")
          gsub (/ /, "#", X)
          sub (/Y/, $1, X)
          gsub (/./, "& ", X)
          for (i=1; i<=L; i+=2) print substr (X, i, L-1)
        }
' $1 

但第一个词只起作用

# # h e l
# h e l l
h e l l o
e l l o #
l l o # #
# # # r e a
# # r e a d
# r e a d e
r e a d e r
e a d e r #
a d e r # #

Answer 1

我会用这样的东西:

awk '{n=length($0)                    # get the length N of the string
      $0 = "##" $0 "##"               # prepend and append "##"
      gsub(/./, "& ")                 # add a space after every character
      for (i=1; i<=2*n; i+=2)         # loop X from position 1 to length of the string
          print substr($0, i, 5*2-1)  # print 5*2 chars from position 2X (-1 for the trailing space)
      print ""}' file                 # print an empty line to separate blocks

查看实际效果:

$ awk '{n=length($0); $0 = "##" $0 "##"; gsub(/./, "& "); {for (i=1; i<=2*n; i+=2) print substr($0, i, 5*2)} print ""}' file
# # h e l
# h e l l
h e l l o
e l l o #
l l o # #

# # r e a
# r e a d
r e a d e
e a d e r
a d e r #
d e r # #

如您所见，这里的关键是硬编码要打印的字符数，而不是依赖于字符串的长度。就我而言，我将其设置为 5。