[SOLVED] What is the syntax to use octal number in xargs for delimiter argument?

tprabhu1983 · 05-11-2017, 04:22 AM

man xargs:
--delimiter=delim, -d
The specified delimiter may be a single character, a C-style character escape such as \n, or an octal or hexadecimal escape code. Octal and hexadecimal escape codes are understood as for the printf command.

The reason is:
Actual Input:
[root@rhel5 ~]# xargs -d\n
No
no
nn

Actual Output:
No
o

Expected Output:
No
no
nn

!!! · 05-11-2017, 06:55 AM

Try -d\\n or -d\\12 or -d\\xa See man printf. I think the shell "eats" the first backslash. Experiment using echo.

hydrurga · 05-11-2017, 06:57 AM

It's treating the \n as a literal 'n'.

Try:

Code:

xargs -d '\n'

tprabhu1983 · 05-11-2017, 09:00 PM

Quote:

Originally Posted by !!!

Try -d\\n or -d\\12 or -d\\xa See man printf. I think the shell "eats" the first backslash. Experiment using echo.

Thanks for your reply.
With your suggestion, character "n" being saved, but it loses "escape character behaviour" of printing on the next line.

[root@rhel5 ~]# xargs -d\\xa
n1
n2
n3
n4

n1 n2 n3 n4
[root@rhel5 ~]# xargs -d\\12
n1
n2
n3

n1 n2 n3
[root@rhel5 ~]# xargs -d\\n
n1
n2
n3

n1 n2 n3
[root@rhel5 ~]# xargs -d\n
n1
n2
n3

1
2
3

tprabhu1983 · 05-11-2017, 09:01 PM

Quote:

Originally Posted by hydrurga

It's treating the \n as a literal 'n'.

Try:

Code:

xargs -d '\n'

Thanks mate. With the below command, it has lost the "escape character behavior"

[root@rhel5 ~]# xargs -d '\n'
n1
n2
n3
n1 n2 n3
[root@rhel5 ~]# echo 1 \
> 2
1 2
[root@rhel5 ~]# echo 1\
> 2\
> 3\
> 4 |xargs -d\n
1234

scasey · 05-12-2017, 11:25 AM

Quote:

Originally Posted by tprabhu1983

Thanks for your reply.
With your suggestion, character "n" being saved, but it loses "escape character behaviour" of printing on the next line.

[root@rhel5 ~]# xargs -d\\xa
n1
n2
n3
n4

n1 n2 n3 n4
[root@rhel5 ~]# xargs -d\\12
n1
n2
n3

n1 n2 n3
[root@rhel5 ~]# xargs -d\\n
n1
n2
n3

n1 n2 n3
[root@rhel5 ~]# xargs -d\n
n1
n2
n3

1
2
3

Typically the defined delimiter is not printed, so yes, it is "lost."

!!! · 05-13-2017, 10:49 AM

Add -n1
Or try -l
See man xargs. The -d seems to be for processing the input; -n for output.
Let us know, and use ThreadTools at top when resolved.

p.s. I meant: echo \n; echo \\n; echo '\n'; n="\n" echo $n

tprabhu1983 · 05-14-2017, 09:19 AM

Quote:

Originally Posted by !!!

Add -n1
Or try -l
See man xargs. The -d seems to be for processing the input; -n for output.
Let us know, and use ThreadTools at top when resolved.

p.s. I meant: echo \n; echo \\n; echo '\n'; n="\n" echo $n

Perfect answer, thanks mate.

[root@rhel5 ~]# echo 1n 2n 3n |xargs -n1
1n
2n
3n

[root@rhel5 ~]# echo 1n 2n 3n |xargs -l
1n 2n 3n