Question

我正在尝试用空行替换两个模式之间的代码块尝试使用以下命令

sed '/PATTERN-1/,/PATTERN-2/d' input.pl

但它只删除了模式之间的界限

PATTERN-1：“= head” PATTERN-2：“= cut”

input.pl包含以下文字

=head
hello
hello world
world
morning
gud
=cut

必需的输出：

=head





=cut

有人可以帮我吗？

Answer 1

$ awk '/=cut/{f=0} {print (f ? "" : $0)} /=head/{f=1}' file
=head





=cut

Answer 2

要修改给定的sed命令，请尝试

$ sed '/=head/,/=cut/{//! s/.*//}' ip.txt
=head





=cut

//!匹配起始/结束范围以外的其他内容，可能取决于sed实现是否动态匹配两个范围或静态匹配其中一个。适用于GNU sed
- s/.*//清除这些行

Answer 3

awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile

# OR
# ^ to take care of line starts with regexp
awk '/^=cut/{found=0}found{print "";next}/^=head/{found=1}1' infile

说明：

awk '/=cut/{                    # if line contains regexp 
        found=0                 # set variable found = 0
     }
     found{                     # if variable found is nonzero value
       print "";                # print ""
       next                     # go to next line
     }
     /=head/{                   # if line contains regexp
       found=1                  # set variable found = 1
     }1                         # 1 at the end does default operation
                                # print current line/row/record
   ' infile

测试结果：

$ cat infile
=head
hello
hello world
world
morning
gud
=cut

$ awk '/=cut/{found=0}found{print "";next}/=head/{found=1}1' infile
=head





=cut

Answer 4

这可能适合你（GNU sed）：

sed '/=head/,/=cut/{//!z}' file

删除=head和=cut之间的行。

如何用空行替换两个模式之间的代码块？

4 个答案: