如何使用包含特定字符串的一个命令在grep regex命令中使用特定字符串跳过注释？

Question

我有这个内容的文件：

path=/nfs/location/alex 
path = /nfs/location/ 
alex/nfs/location 
/nfs/location/
# /nfs/location/
#alex /nfs/location
alex # d /nfs/location
/nfs/location

我想打印所有没有评论＃的行，并包含以/ nfs / location

开头的路径

path=/nfs/location/alex 
path = /nfs/location/ 
/nfs/location/
/nfs/location

我写了unix commad：

grep -rE ([^#.*]|^[^0-9aA-zZ]/nfs/location/ .

但带＃的行也会出现

Answer 1

GNU grep 方法：

grep -P '^([^#]+[[:space:]]*=[[:space:]]*)?/nfs/location' file

输出：

path=/nfs/location/alex 
path = /nfs/location/ 
/nfs/location/
/nfs/location

Answer 2

只需使用awk：

awk -F'[= ]+' 'index($NF,"/nfs/location")==1 && !/#/' file

Answer 3

awk '/^path|^\/nfs/' file

path=/nfs/location/alex 
path = /nfs/location/ 
/nfs/location/
/nfs/location