从文件夹中的多个文件中提取类似行

使用sed,您可以使用地址范围输出行块:

sed -n '/^[[:blank:]]*class[[:blank:]]/,/IS_CRITICAL/p' file.py

编辑:

补充 [[:blank:]] 前后 class 仅匹配前面有零个或多个空格或制表符的类定义。

试试这个,看看结果是否符合您的要求(GNU awk):

awk '/IS_CRITICAL/{sub(/IS_CRITICAL.*/,"IS_CRITICAL");print "class " $0}' RS="class " all.py

使用Perl一行程序

 perl -0777 -ne ' while( /(\bclass\s*.+?IS_CRITICAL)/gs ) { print "$1\n" } ' 

输入:

$ cat josh.py
import stuff

class BarFoo001(BarFooBase):

    info = self.info
    description = 'here's the stuff I want'
    IS_CRITICAL = true

    def method(sdf):
        etc...
    def method2(fddf):
        print
$ perl -0777 -ne ' while( /(\bclass\s*.+?IS_CRITICAL)/gs ) { print "$1\n" } ' josh.py
class BarFoo001(BarFooBase):

    info = self.info
    description = 'here's the stuff I want'
    IS_CRITICAL
$

perl -0777 -ne ' while( /(\bclass\s*.+?IS_CRITICAL)/gs ) { print "$ARGV:$1\n" } ' *py

awk 但这些版本无法在所有版本或不同的O.S系统中测试。

awk '
{
  sub(/^ +/,"")
}
/class/{
  found=1
}
/IS_CRITICAL/ && found{
  sub(/ =.*/,"")
  print
  found=""
}
found
'  Input_file

$ grep -E '^[[:space:]]*(class|description)[[:space:]]' file
class BarFoo001(BarFooBase):
    description = 'here's the stuff I want'

$ awk 'sub(/^[[:space:]]*(class|description =)[[:space:]]+/,"")' file
BarFoo001(BarFooBase):
'here's the stuff I want'