Linux piping data containing backticks - linux

I'm piping output from one command into a second:
mpc listall | mpc add
mpc listall returns the following data (can output 1 or more lines):
Dare - 16 - I´ll Be Your King.mp3
When piping it into the next command, it seems that my shell (Ash on BusyBox) converts the ´ into an asterisk, as I get the error
error adding Dare - 16 - I*ll Be Your King.mp3: No such directory
Manually adding double quotes works! like this:
mpc add "Dare - 16 - I´ll Be Your King.mp3"
So, I tried adding those by sed and awk, but in those cases the backtick gets converted to an asterisk again:
mpc listall | sed 's/^/"/;s/$/"/'
returns:
"Dare - 16 - I*ll Be Your King.mp3"
So, question is, is there a way to pass backticks, or actually any character, as is, into another command without conversions?
By the way, obviously, it's quite bad to have this character in the file name in the first place, but I want my code to be robust and able to handle anything thrown at it.

Not a direct solution the problem, but a work-around (that's actually better, because it uses the API) in python using the python-mpd2 library:
myMpdClient = MPDClient()
myMpdClient.connect("localhost", 6600) # connect to localhost:6600
myMpdClient.findadd('base','directory-name')
myMpdClient.close()
I'll accept a working solution to the actual proposed problem, instead of this one, if someone posts a working solution.

Related

sed behavior: Pattern space and Address range [duplicate]

I noticed something a bit odd while fooling around with sed. If you try to remove multiple line intervals (by number) from a file, but any interval specified later in the list is fully contained within an interval earlier in the list, then an additional single line is removed after the specified (larger) interval.
seq 10 > foo.txt
sed '2,7d;3,6d' foo.txt
1
9
10
This behaviour was behind an annoying bug for me, since in my script I generated the interval endpoints on the fly, and in some cases the intervals produced were redundant. I can clean this up, but I can't think of a good reason why sed would behave this way on purpose.
Since this question was highlighted as needing an answer in the Stack Overflow Weekly Newsletter email for 2015-02-24, I'm converting the comments above (which provide the answer) into a formal answer. Unattributed comments here were made by me in essentially equivalent form.
Thank you for a concise, complete question. The result is interesting. I can reproduce it with your script. Intriguingly, sed '3,6d;2,7d' foo.txt (with the delete operations in the reverse order) produces the expected answer with 8 included in the output. That makes it look like it might be a reportable bug in (GNU) sed, especially as BSD sed (on Mac OS X 10.10.2 Yosemite) works correctly with the operations in either order. I tested using 'sed (GNU sed) 4.2.2' from an Ubuntu 14.04 derivative.
More data points for you/them. Both of these include 8 in the output:
sed -e '/2/,/7/d' -e '/3/,/6/d' foo.txt
sed -e '2,7d' -e '/3/,/6/d' foo.txt
By contrast, this does not:
sed -e '/2/,/7/d' -e '3,6d' foo.txt
The latter surprised me (even accepting the basic bug).
Beats me. I thought given some of sed's arcane constructs that you might be missing the batman symbol or something from the middle of your command but sed -e '2,7d' -e '3,6d' foo.txt behaves the same way and swapping the order produces the expected results (GNU sed 4.2.2 on Cygwin). /bin/sed on Solaris always produces the expected result and interestingly so does GNU sed 3.02. Ed Morton
More data: it only seems to happen with sed 4.2.2 if the 2nd range is a subset of the first: sed '2,5d;2,5d' shows the bug, sed '2,5d;1,5d' and sed '2,5d;2,6d' do not. glenn jackman
The GNU sed home page says "Please send bug reports to bug-sed at gnu.org" (except it has an # in place of ' at '). You've got a good reproduction; be explicit about the output you expect vs the output you get (they'll get the point, but it's best to make sure they can't misunderstand). Point out that the reverse ordering of the commands works as expected, and give the various other commands as examples of working or not working. (You could even give this Q&A URL as a cross-reference, but make sure that the bug report is self-contained so that it can be understood even if no-one follows the URL.)
You can also point to BSD sed (and the Solaris version, and the older GNU 3.02 sed) as behaving as expected. With the old version GNU sed working, it means this is arguably a regression. […After a little experimentation…] The breakage occurred in the 4.1 release; the 4.0.9 release is OK. (I also checked 4.1.5 and 4.2.1; both are broken.) That will help the maintainers if they want to find the trouble by looking at what changed.
The OP noted:
Thanks everyone for comments and additional tests. I'll submit a bug report to GNU sed and post their response. santayana

Read filename with * shell bash

I'am new in Linux and I want to write a bash script that can read in a file name of a directory that starts with LED + some numbers.(Ex.: LED5.5.002)
In that directory there is only one file that will starts with LED. The problem is that this file will every time be updated, so the next time it will be for example LED6.5.012 and counting.
I searched and tried a little bit and came to this solution:
export fspec=/home/led/LED*
LedV=`basename $fspec`
echo $LedV
If I give in those commands one by one in my terminal it works fine, LedV= LED5.5.002 but if i run it in a bash scripts it gives the result: LedV = LED*
I search after another solution:
a=/home/led/LED*
LedV=$(basename $a)
echo $LedV
but here again the same, if i give it in one by one it's ok but in a script: LedV = LED*.
It's probably something small but because of my lack of knowledge over Linux I cannot find it. So can someone tell what is wrong?
Thanks! Jan
Shell expansions don't happen on scalar assignments, so in
varname=foo*
the expansion of "$varname" will literally be "foo*". It's more confusing when you consider that echo $varname (or in your case basename $varname; either way without the double quotes) will cause the expansion itself to be treated as a glob, so you may well think the variable contains all those filenames.
Array expansions are another story. You might just want
fspec=( /path/LED* )
echo "${fspec[0]##*/}" # A parameter expansion to strip off the dirname
That will work fine for bash. Since POSIX sh doesn't have arrays like this, I like to give an alternative approach:
for fspec in /path/LED*; do
break
done
echo "${fspec##*/}"
pwd
/usr/local/src
ls -1 /usr/local/src/mysql*
/usr/local/src/mysql-cluster-gpl-7.3.4-linux-glibc2.5-x86_64.tar.gz
/usr/local/src/mysql-dump_test_all_dbs.sql
if you only have 1 file, you will only get 1 result
MyFile=`ls -1 /home/led/LED*`

Understanding sed

I am trying to understand how
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
worked and what the pieces mean. The man page I read just confused me more and I tried the info sai Id but had no idea how to work it! I'm pretty new to Linux. Debian is my first distro but seemed like a rather logical place to start as it is a root of many others and has been around a while so probably is doing stuff well and fairly standardized. I am running Wheezy 64 bit as fyi if needed.
The sed command is a stream editor, reading its file (or STDIN) for input, applying commands to the input, and presenting the results (if any) to the output (STDOUT).
The general syntax for sed is
sed [OPTIONS] COMMAND FILE
In the shell command you gave:
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
the sed command is s/\^\[/\o33/g;s/\[1G\[/\[27G\[/' and /var/log/boot is the file.
The given sed command is actually two separate commands:
s/\^\[/\o33/g
s/\[1G\[/\[27G\[/
The intent of #1, the s (substitute) command, is to replace all occurrences of '^[' with an octal value of 033 (the ESC character). However, there is a mistake in this sed command. The proper bash syntax for an escaped octal code is \nnn, so the proper way for this sed command to have been written is:
s/\^\[/\033/g
Notice the trailing g after the replacement string? It means to perform a global replacement; without it, only the first occurrence would be changed.
The purpose of #2 is to replace all occurrences of the string \[1G\[ with \[27G\[. However, this command also has a mistake: a trailing g is needed to cause a global replacement. So, this second command needs to be written like this:
s/\[1G\[/\[27G\[/g
Finally, putting all this together, the two sed commands are applied across the contents of the /var/log/boot file, where the output has had all occurrences of ^[ converted into \033, and the strings \[1G\[ have been converted to \[27G\[.

How do I print the first few lines of a makefile (-Wall -g flags)?

How do I get the first n lines of the output of a makefile (specifically, my complier is g++). Either a script in linux or in the makefile would work (if you could provide both, that'll be even better).
I have tried
make | head -n 5
but it's not working.
Currently, the process I go through is tedious; I'm piping the output to a text file before using head on it (then having to delete the file).
Given that the messages from the compiler appear on standard error rather than standard output, you need to redirect both:
make 2>&1 | head -n 20
(I think 5 lines will be too small to be useful.)

change multiple files commandline

I have separated some tracks from mp3 mixes using mp3splt.
BASH: (mp3splt -c('**!!***use .cue file***!!**') [cuefile.cue] [nonstopmix.mp3] ~for anyone interested, is in the Ubu repos~)
And I ended up with these filenames: "Antares" - 01 - "Xibalba".mp3 which is not a format I prefer, now I've made it a little project to change them with a shell script but its more difficult than I anticipated.
I want to change the filename from:
"Antares" - 01 - "Xibalba".mp
to:
01-Antares_-_Xibalba.mp3
so far I've used :
for var in *.mp3; do mv $var {var/"/}; done
and I could repeat that until I'm through, delete the 0x number and add one but I'd like to do it more efficient.
Could anyone give me a pointer (!not a script!) ?
I'd still like to write it myself but there's so much options that I'm a bit lost.
so far I thought to use this program flow:
read all the filenames containing .mp3 and declare as variable $var
strip $var from quotes
select 0x number, append delimiter _ (0x_)
move 0x_ to the beginning of the string
select remaining ' - - ' and change to '-'
done
which bash programs to use? especially changing the 0x puzzles me cuz I need a loop which increments this number and test if it is present in the filename variable and then it has to be changed.
It is easy to do in python 2.x. You can use this logic in any language you want.
import string
a=raw_input('Enter the name of song')
a=a.replace('"', "")
a=a.replace('.mp', ' .mp3')
words = a.split()
print words[2]+'-'+words[0]+'_-_'+words[4]+words[5]
Logic:
I removed ", then make .mp to .mp3, then splitted the string, which created a list ( array ) and then printed the elements according to need.
Try doing this :
rename -n 's/"(\w+)"\s+-\s*(\d+)\s*-\s*"(\w+)"\.mp/$2-$1_-_$3.mp3/' *mp
from the shell prompt. It's very useful, you can put some perl tricks like I does in a substitution.
You can remove the -n (dry-run mode switch) when your tests become valids.
There are other tools with the same name which may or may not be able to do this, so be careful.
If you run the following command (linux)
$ file $(readlink -f $(type -p rename))
and you have a result like
.../rename: Perl script, ASCII text executable
then this seems to be the right tool =)
If not, to make it the default (usually already the case) on Debian and derivative like Ubuntu :
$ sudo update-alternatives --set rename /path/to/rename
Last but not least, this tool was originally written by Larry Wall, the Perl's dad.

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