Hi i would like to do the following.
./script.sh some.file.name.dat another.file.dat
Filename1=$(echo "$1"|cut -d '.' -f 1,2)
Filename2=$(echo "$2"|cut -d '.' -f 1,2)
tempfile_"$1"=$(mktemp)
tempfile_"$2"=$(mktemp)
I know this code isn't working. I need to create these temporary files and use them in a for loop later, in which i will do something with the input files and save the output in these temporary files for later usage. So basically i would like to create the variable names dependent on the name of my input files.
I googled a lot and didn't found any answers to my problem.
I would like to thank for your suggestions
I'll suggest an alternate solution to use instead of entering the variable naming hell you're proposing (Using the variables later will cause you the same problems later, the scale will just be magnified).
Use Associative arrays (like tempfile[$filename]) instead of tempfile_"$filename". That's what associative arrays are for:
declare -A tempfile
tempfile[$1]=$(mktemp)
tempfile[$2]=$(mktemp)
cat ${tempfile[$1]}
cat ${tempfile[$2]}
rm -f ${tempfile[$1]}
rm -f ${tempfile[$2]}
Note: Associative arrays require bash version 4.0.0 or newer.
If you dont have Bash version 4.0.0 or newer, see the following answers for great workarounds that dont use eval.
How to define hash tables in Bash?
Associative arrays in Shell scripts
You cannot do that since Bash just doesn't allow dots in the names of identifiers/variables. Both of your arguments $1 and $2 have dots (periods) in them and that cannot be used in variable names you're trying to create i.e. tempfile_$1
See this page for details.
Try this:
eval "tempfile_"$1"=$(mktemp)"
eval "tempfile_"$2"=$(mktemp)"
Use something like:
f1=`echo $1|tr '.' '_'`
declare "tempfile_$f1=$(mktemp)"
Check out How to define hash tables in bash?
Related
How can I pass each one of my repository files and to do something with them?
For instance, I want to make a script:
#!/bin/bash
cd /myself
#for-loop that will select one by one all the files in /myself
#for each X file I will do this:
tar -cvfz X.tar.gz /myself2
So a for loop in bash is similar to python's model (or maybe the other way around?).
The model goes "for instance in list":
for some_instance in "${MY_ARRAY[#]}"; do
echo "doing something with $some_instance"
done
To get a list of files in a directory, the quick and dirty way is to parse the output of ls and slurp it into an array, a-la array=($(ls))
To quick explain what's going on here to the best of my knowledge, assigning a variable to a space-delimited string surrounded with parens splits the string and turns it into a list.
Downside of parsing ls is that it doesn't take into account files with spaces in their names. For that, I'll leave you with a link to turning a directory's contents into an array, the same place I lovingly :) ripped off the original array=($(ls -d */)) command.
you can use while loop, as it will take care of whole lines that include spaces as well:
#!/bin/bash
cd /myself
ls|while read f
do
tar -cvfz "$f.tar.gz" "$f"
done
you can try this way also.
for i in $(ls /myself/*)
do
tar -cvfz $f.tar.gz /myfile2
done
I want to store output of ls command in my bash script in a variable and use each file name in a loop, but for example one file in the directory has name "Hello world", when I do variable=$(ls) "Hello" and "world" end up as two separate entries, and when I try to do
for i in $variable
do
mv $i ~
done
it shows error that files "Hello" and "world" doesn't exist.
Is there any way I can access all files in current directory and run some command even if the files have space(s) in their names.
If you must, dirfiles=(/path/of/interest/*).
And accept the admonition against parsing the output of ls!
I understand you are new to this and I'd like to help. But it isn't easy for me (us?) to provide you with an answer that would be of much help to you by the way you've stated your question.
Based on what I hear so far, you don't seem to have a basic understanding on how parameter expansions work in the shell. The following two links will be useful to you:
Matching Pathnames, Parameters
Now, if your task at hand is to operate on files meeting certain criteria then find(1) will likely to do the job.
Say it with me: don't parse the output of ls! For more information, see this post on Unix.SE.
A better way of doing this is:
for i in *
do
mv -- "$i" ~
done
or simply
mv -- * ~
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*`
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.
I am experimenting with wildcards in bash and tried to list all the files that start with "xyz" but does not end with ".TXT" but getting incorrect results.
Here is the command that I tried:
$ ls -l xyz*[!\.TXT]
It is not listing the files with names "xyz" and "xyzTXT" that I have in my directory. However, it lists "xyz1", "xyz123".
It seems like adding [!\.TXT] after "xyz*" made the shell look for something that start with "xyz" and has at least one character after it.
Any ideas why it is happening and how to correct this command? I know it can be achieved using other commands but I am especially interested in knowing why it is failing and if it can done just using wildcards.
These commands will do what you want
shopt -s extglob
ls -l xyz!(*.TXT)
shopt -u extglob
The reason why your command doesn't work is beacause xyz*[!\.TXT] which is equivalent to xyz*[!\.TX] means xyz followed by any sequence of character (*) and finally a character in set {!,\,.,T,X} so matches 'xyzwhateveryouwant!' 'xyzwhateveryouwant\' 'xyzwhateveryouwant.' 'xyzwhateveryouwantT' 'xyzwhateveryouwantX'
EDIT: where whateveryouwant does not contain any of !\.TX
I don't think this is doable with only wildcards.
Your command isn't working because it means:
Match everything that has xyz followed by whatever you want and it must not end with sequent character: \, .,T and X. The second T doesn't count as far as what you have inside [] is read as a family of character and not as a string as you thought.
You don't either need to 'escape' . as long as it has no special meaning inside a wildcard.
At least, this is my knowledge of wildcards.