bash question about sed '"$i"!d' listofusers

haliban · 09-21-2008, 06:46 AM

Hello All,

Here it is a part of my script.

Code:

#!/bin/bash

users=/web/users
max=`/bin/cut -f1 users | wc -l`


        for ((i=1; i <= $max ; i++))
        do
         curusr= sed '"$i"!d' users
                                              
 done

I have list of users in file plain text file named "users" and on each line there is just one username.

- with

Code:

 max=`/bin/cut -f1 users | wc -l`

I am getting the whole number of lines/users of the file and use it in "for" loop.

- When I try to

Code:

curusr= sed '"$i"!d' users

get actual username from each line I get an error message :

Code:

+ sed '"$i"!d' users
sed: -e expression #1, char 1: unknown command: `"'

When I execute

Code:

 sed '2!d' users

from the console I get the username from the second line.

I assume there is something wrong in my syntax. I look around the Internet and found a lot of interesting topics but not how to fix it.

Any help how to make it working ?

Thank you for your help!

ghostdog74 · 09-21-2008, 06:51 AM

how does your input file look like and what is your final output?

pixellany · 09-21-2008, 07:56 AM

It looks like you are using addressing, which has the following syntax:

sed '<line#><action>'
OR
sed '/<pattern>/<action>'

Also, the quoting needs to be changed: single-quotes protect everything from BASH, so, the double-quote is being sent to SED.

I think you need:
sed "/$i/!d"

But, why the negation? The same result can be gotten this way:
sed -n "/$1/p"

Excellent SED tutorial here:
http://www.grymoire.com/Unix/Sed.html

haliban · 09-24-2008, 02:01 AM

Thank you for your help guys ! :-)

pixellany, I have read in SED tutorial that there are 3 different ways to print specific line from text file and chose
sed "/$i/!d"

I put in my script sed -n "/$1/p" as you advised.

There is something very interesting I have noticed, when I have

Code:

sed -n "/1/p" users

inside the script it shows me the user name located on the first line, but if I do

Code:

sed -n "/3/p" users

It does not show me anything (no error message). On the other hand if I do

Code:

sed -n '3 p' users

from shell it gives me the user name on line 3 :-) Funny....

The file with users looks like :

Code:

[myuser@theserver webusers]$ cat users 
user1
user2
user3
user4
...

...
...
user35

I feel that there is something small and silly I am missing .... do not know !

Sed tutorial is very valuable, thanks again ;-)

Mr. C. · 09-24-2008, 02:12 AM

I get the correct results:

Code:

$ sed -n "/3/p" users
user3
$ sed -n "3p" users  
user3

Take note to what Pixellany indicated about the differences between an input line number and a line that contains a pattern.

nitin_nitt · 09-25-2008, 03:14 AM

Try this:

Code:

for i in `cut -f 1 /web/users`; do curuser=echo $i; done

If you don't want to change the code for some reason, change the sed command as:

Code:

sed -n $i'p' users

haliban · 09-25-2008, 06:42 PM

Thank you very much for your help !

All good and working so far ;-)