find out password is *

fzhcl

Hi  Guys,

Boss ask me to find out which account's password is *,
below is the content of passwd file This is AIX unix file)

root:
        password = HWjXEGfGcjl2c
        lastupdate = 1329402022

daemon:
        password = *

bin:
        password = *

sys:
        password = *

adm:
        password = 70AjL2hqdkVoI
        lastupdate = 1288287263
        flags = ADMCHG

uucp:
        password = *

guest:
        password = *


I need the following output:

daemon:
bin:
sys:
uucp:
guest:


Can anyone help me out?  Thanks in advance.

rdcwayx

1. awk 'BEGIN{RS="";FS="\n"} /password = \*/{print $1}' infile

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waker

1. 
   
2. echo 'g/password = [*]/-1p'|ed -s urfile
   

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yangkyo821

1. sed -n '/password = \*/{g;p};h' file

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zhaopingzi

1. 
   
2. $   awk -vRS="" -vFS="\n"  '{if($2=="        password = *")print $1}' passwdfile
   
3. daemon:
   
4. bin:
   
5. sys:
   
6. uucp:
   
7. guest:
   

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1. 
   
2. $   awk -vRS="" -vFS="\n"  '{if($2~/password = \*/)print $1}' passwdfile
   
3. daemon:
   
4. bin:
   
5. sys:
   
6. uucp:
   
7. guest:
   

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ajdan4q

1. sed -n '/:/{h;n};/*/{x;p}' pw

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fzhcl

awk 'BEGIN{RS="";FS="\n"} /password = \*/{print $1}'  
echo 'g/password = [*]/-1p'|ed -s urfile  
awk -vRS="" -vFS="\n"  '{if($2=="        password = *")print $1}' passwdfile
awk -vRS="" -vFS="\n"  '{if($2~/password = \*/)print $1}'   

the above works very well in AIX.

The following only works in Linux,

sed -n '/:/{h;n};/*/{x;p}'  
sed -n '/password = \*/{g;p};h' file
sed -rn '/^[^ \t]+/{N;/password = \*/P}'
      -r, --regexp-extended   use extended regular expressions in the script
      AIX doesn't have -r option but do have regexp command.
     I will figure out how regexp  work in the future.

I really appreciate your help.   Thanks guys.

fzhcl

Finally understand the RS="" meaning

The empty string, "" (a string of no characters), has a special meaning as the value of RS: it means that records are separated only by blank lines

mpstat

回复 3# waker


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