关于awk分隔符FS的一个疑问

bikkuri

大家好！
我有一个问题向大家请教。
有一个字符串/home/box/.z/.upgrade，我想得到两个/.之间的z，也就是以/.为分隔符，但是awk会把.正则匹配为任意字符。
[box@5f72b64c650c:~/.z/.upgrade]$ pwd
/home/box/.z/.upgrade
[box@5f72b64c650c:~/.z/.upgrade]$ pwd|awk -F"/." '{print$2}'
ome
[box@5f72b64c650c:~/.z/.upgrade]$ pwd|awk -F'/.' '{print$2}'
ome
[box@5f72b64c650c:~/.z/.upgrade]$ pwd|awk -F"/\." '{print$2}'
awk: warning: escape sequence `\.' treated as plain `.'
ome
[box@5f72b64c650c:~/.z/.upgrade]$ pwd|awk -F'[/.]+' '{print$2}'
home
[box@5f72b64c650c:~/.z/.upgrade]$ pwd|grep -Po "(?<=/\.).*(?=/\.)"
z
[box@5f72b64c650c:~/.z/.upgrade]$
请问如果用awk的话应该怎么设置分隔符FS呢？

moperyblue

1. echo '/home/box/.z/.upgrade'|awk -F'/\\.' '{print $2}'
   
2. z
   
3. echo '/home/box/.z/.upgrade'|awk -F'/[.]' '{print $2}'
   
4. z
   
5. echo '/home/box/.z/.upgrade'|awk -F/[.] '{print $2}'  
   
6. z
   
7. echo '/home/box/.z/.upgrade'|awk -F"/[.]" '{print $2}'
   
8. z
   
9. 
   
10. echo '/home/box/.z/.upgrade'|sed '/\n/P;s#/\.#\n#g;D'
    
11. z
    
12. echo '/home/box/.z/.upgrade'|sed 'ta;s#/\.#\n#;D;:a;s#/\..*##'
    
13. z

复制代码

hz_oracle

转译啊，\.  

liion631818

回复 1# bikkuri
不同版本还是有点差异的

li@ubuntu:~/shell/python$ echo '/home/box/.z/.upgrade' | awk -F"/\." '{print$2}'
z

li@ubuntu:~/shell/python$ echo '/home/box/.z/.upgrade' | awk -F"/\\." '{print$2}'
z

li@ubuntu:~/shell/python$ echo '/home/box/.z/.upgrade' | awk -F"/\\\." '{print$2}'
z


li@ubuntu:~/shell/python$ echo '/home/box/.z/.upgrade' | awk -F'/\.' '{print$2}'
z

li@ubuntu:~/shell/python$ echo '/home/box/.z/.upgrade' | awk -F'/\\.' '{print$2}'
z


   

moperyblue

1. .

复制代码

jason680

回复 1# bikkuri

try this way ...

$ echo '/home/box/.z/.upgrade' | awk -F'/[.]' '{print $2}'
z

   

blackold

-F '/[.]'  或 -F '/\\.'

jason680

(网络问题，重发.....)

   

jason680

(网络问题，重发.....)