Pass variable as insert with SED. 3

[AnotherAlan]

Hi all,

Is it possible to insert more than one string with sed.

i.e Something like;
sed '$i\
`cat /tmp/test/ng_pass` ' /tmp/test/pass_test > /tmp/test/pass_test.updated

I would like to insert the contents of a file / variable.
In this case the file ng_pass contains 2 lines;
UK
US
I have dynamic entries and the contents of ng_pass may vary from 1 to 4 different lines that need inserting.
At the moment I have tried with files and variables but cannot get the syntax correct.

Any help appreciated.
Thanks
Alan

 

[feherke]

Hi

awk 'FNR==NR{d[NR]=$0;next}/insert condition/{for(i=1;i in d;i++)print d[i]}{print}' /insertable/file /input/file

Feherke.

http://rootshell.be/~feherke/

 

[AnotherAlan]

Hi Feherke,

Thanks for the reply and the solution.
I have a small query.
The file ng_pass is built on the fly after certain conditions have been met earlier in the script and will therefore always apply when it gets to this part.

I am confused as to this part of your code:
/insert condition/ - and how to make it work for me.

Thanks again.

 

[feherke]

Hi

Oops, I misread your code. So you want the second file before the last line.

awk 'NR>1{print s}{s=$0}END{while(getline<"/insertable/file")print;print s}' /input/file

Feherke.

http://rootshell.be/~feherke/

 

[AnotherAlan]

Hi Feherke,

I apologise for keeping you busy.
This doesn't seem to do what I was expecting.

I think I may have explained my problem poorly.
I am automating the roll-out of NIS.
To do this I need to build a list of groups who are allowed access. This list is built as a file, ng_pass, and will contain entries such as admin, support, tsg e.t.c

I would like to then insert, before the last line, each item in the file as a seperate line in the password and shadow files.

The last line of the file is +:x:::::/bin/false

I would like all new entries to precede this line;

Expected output would be;
admin
support
tsg
+:x:::::/bin/false

I can workaround this problem by adding the last line (which never changes) to the ng_pass file, deleting the last line form the password file and just append.
I was trying to be a little too clever for my own good.

Thanks for your help with this, I do appreciate it.
Alan

 

[feherke]

Hi

Then let us take some example :

[blue]master #[/blue] cat [green][i]pass_test[/i][/green]
1
2
3
4
last

[blue]master #[/blue] cat [green][i]ng_pass[/i][/green]
one
two
three

[blue]master #[/blue] awk 'NR>1{print s}{s=$0}END{while(getline<"[green][i]ng_pass[/i][/green]")print;print s}' [green][i]pass_test[/i][/green]
1
2
3
4
one
two
three
last

The content of ng_pass is before the last line of pass_test. Where am I wrong ?

Feherke.

http://rootshell.be/~feherke/

 

[AnotherAlan]

Thanks Feherke.

You are most definately NOT wrong.
I am running Solaris9, changed awk for nawk and good to go.

Much appreciated as always.

Alan

 

[PHV]

Unless you like the possibility of endless loop, replace this:
while(getline<"ng_pass")
with this:
while((getline<"ng_pass")>1)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[feherke]

Hi

Ahh. Then was the [tt]getline[/tt]. I should copy a couple of times PHV's portable solutions. He had a comment on such issue just a few days ago...

Feherke.

http://rootshell.be/~feherke/

 

[feherke]

Hi

Hmm... PHV's hand was faster then my mind... Thank you.

Feherke.

http://rootshell.be/~feherke/

 

[AnotherAlan]

Thank you Gentlemen.

One day I will get to grips with this awk /sed syntax.
I've read the o'reilly book, cover to cover, but it just doesn't stick.

Anyway, appreciated as always.
Alan

 

[bigoldbulldog]

For a sed/subshell solution I found:

sed "\$i\\
$(sed '$!s/$/\\/' ng_pass)
" pass_test

Cheers,
ND

 

[AnotherAlan]

Nice work ND, works like a dream.

Thanks for the effort.
Alan