Re: [Help-bash] how to understand the xargs

[lina]

On Tuesday 27,December,2011 11:02 PM, Greg Wooledge wrote:
> On Tue, Dec 27, 2011 at 12:11:59AM +0800, lina wrote:
> > I don't understand the xargs well,
> There's really only one thing you need to know about xargs: never use it.
>
> (Well, OK, I exaggerate very slightly.  It's OK to use it if you supply
> an option that suppresses its native uselessness, such as GNU's -0.)
>
> > can someone help me understand the below sentence,
> >
> > ls . | xargs -i -t cp ./{} {}.bak
> Never use the output of ls in a script.  Ever.
>
> If your goal here is to make a backup of each file in the current directory,
> then do this instead:
>
> for f in ./*; do cp "$f" "$f.bak"; done
>
> Using a glob (like ./*) instead of the output of ls is always better.
> Globs are not mangled by ls (replacing non-printable characters with
> question marks, failing completely on files with newlines in their names,
> etc.), and because you are not using xargs, you will also not have any
> of the problems that xargs has with filenames (spaces, quotes).
>
> > How does xargs work,
> Badly.
>
> imadev:~$ echo 'hi"mom' | xargs echo
> xargs: missing quote?: himom
>
> imadev:~$ echo "it doesn't work" | xargs echo
> xargs: missing quote?: doesnt work
>
> imadev:~$ echo "this is one filename.txt" | xargs ./bin/args rm
> 5 args:<rm>  <this>  <is>  <one>  <filename.txt>
>
> > ls | xargs -i cp ./{} {basename {}.bak}
> > definitely not work,
> I don't know what that example is supposed to do.  You realize that the
> basename command is used to *strip out* directory paths from path names,
> or suffixes from file names?  (Also, that is not how you do a command
> substitution... but since I can't tell what you were trying to do, I can't
> write a working example.)
>
>
> On Tue, Dec 27, 2011 at 12:25:52AM +0800, lina wrote:
> > is the help-bash an active list?
> >
> > seems no reply.
> I just got your messages a few minutes ago.

Thanks for your reply.

another question,
$ for i in {1..10} ; do echo $i ; done
1
2
3
4
5
6
7
8
9
10

in command line it works, while on

$ cat a.sh
#!/bin/sh

for i in {1..10} ; do

echo $i

done
$ ./a.sh
{1..10}

why put in a script, it does not work.

BTW, on script I used to use for i in $(seq 10)
just recent trying to learn more, so do some practice.

I will avoid using xargs,

Thanks with best regards,