There are 3 txt files called
1.txt 2.txt 3.txt
I want to batch copy with the name
1.txt.cp 2.txt.cp 3.txt.cp
using the wildcard *
I entered the command cp *.txt *.txt.cp
but it wasn't working...
cp : target *.txt.cp : is not a directory
what was the problem???
Use: for i in *.txt; do cp "$i" "$i.cp"; done
Example:
$ ls -l *.txt
-rw-r--r-- 1 halley halley 20 out 27 08:14 1.txt
-rw-r--r-- 1 halley halley 25 out 27 08:14 2.txt
-rw-r--r-- 1 halley halley 33 out 27 08:15 3.txt
$ ls -l *.cp
ls: could not access '*.cp': File or directory does not exist
$ for i in *.txt; do cp "$i" "$i.cp"; done
$ ls -l *.cp
-rw-r--r-- 1 halley halley 20 out 27 08:32 1.txt.cp
-rw-r--r-- 1 halley halley 25 out 27 08:32 2.txt.cp
-rw-r--r-- 1 halley halley 33 out 27 08:32 3.txt.cp
$ for i in *.txt; do diff "$i" "$i.cp"; done
$
If you are used to MS/Windown CMD shell, it is important to note that Unix system handle very differently the wild cards. MS/Windows has kept the MS/DOS rule that said that wild cards were not interpreted but were passed to the command. The command sees the wildcard characters and can handle the second * in the command as noting where the match from the first should go, making copy ab.* cd.* sensible.
In Unix (and derivatives like Linux) the shell is in charge of handling the wildcards and it replaces any word containing one with all the possible matches. The good news is that the command has not to care about that. But the downside is that if the current folder contains ab.txt ab.md5 cd.jpg, a command copy ab.* cd.* will be translated into copy ab.txt ab.md5 cd.jpg which is probably not want you would expect...
The underlying reason is Unix shells are much more versatile than the good old MS/DOS inherited CMD.EXE and do have simple to use for and if compound commands. Just look at #Halley Oliveira's answer for the syntax for your use case.
Related
With sed I try to replace the value 0.1.233... On the command line there is no problem; however, when putting this command in a shell script, I get an error:
sed: couldn't open temporary file ../project/cas-dp-ap/sedwi3jVw: Permission denied
I don't understand where this temporary sedwi file comes from.
Do you have any idea why I have this temporary file and how I can pass it?
$(sed -i "s/$current_version/$version/" $PATHPROJET$CREATE_PACKAGE/Chart.yaml)
++ sed -i s/0.1.233/0.1.234/ ../project/cas-dp-ap/Chart.yaml
sed: couldn't open temporary file ../project/cas-dp-ap/sedwi3jVw: Permission denied
+ printf 'The version has been updated to : 0.1.234 \n\n \n\n'
The version has been updated to : 0.1.234
+ printf '***********************************'
sed -i is "in-place editing". However "in-place" isn't really. What happens is more like:
create a temporary file
run sed on original file and put changes into temporary file
delete original file
rename temporary file as original
For example, if we look at the inode of an edited file we can see that it is changed after sed has run:
$ echo hello > a
$ ln a b
$ ls -lai a b
19005916 -rw-rw-r-- 2 jhnc jhnc 6 Jan 31 12:25 a
19005916 -rw-rw-r-- 2 jhnc jhnc 6 Jan 31 12:25 b
$ sed -i 's/hello/goodbye/' a
$ ls -lai a b
19005942 -rw-rw-r-- 1 jhnc jhnc 8 Jan 31 12:25 a
19005916 -rw-rw-r-- 1 jhnc jhnc 6 Jan 31 12:25 b
$
This means that your script has to be able to create files in the folder where it is doing the "in-place" edit.
The proper syntax is identical on the command line and in a script. If you used $(...) at the prompt then you would have received the same error.
sed -i "s/$current_version/$version/" "$PATHPROJET$CREATE_PACKAGE/Chart.yaml"
(Notice also the quoting around the file name. Probably your private variables should use lower case.)
The syntax
$(command)
takes the output from command and tries to execute it as a command. Usually you would use this construct -- called a command substitution -- to interpolate the output of a command into a string, like
echo "Today is $(date)"
(though date +"Today is %c" is probably a better way to do that particular thing).
Is it possible?
For example, say I want to run the command ll:
My output would look something like this:
josh#zeitgeist ~ ll
total 41148
drwxr-xr-x 42 josh josh 4096 Aug 4 22:52 ./
drwxr-xr-x 4 root root 4096 Jul 9 21:18 ../
-rw-rw-r-- 1 josh josh 3523718 Jul 11 00:17 2017-07-11-001710_3840x2160_scrot.png
but I want it to look like this:
josh#zeitgeist ~ ll
XXXtotal 41148
XXXdrwxr-xr-x 42 josh josh 4096 Aug 4 22:52 ./
XXXdrwxr-xr-x 4 root root 4096 Jul 9 21:18 ../
XXX-rw-rw-r-- 1 josh josh 3523718 Jul 11 00:17 2017-07-11-XXX001710_3840x2160_scrot.png
I already know about using PS1='XXX' to change the prompt; is there a way to change every line of the output that gets displayed, specifically in the terminal(not changing the output and putting it in a file)?
I would like to do this to have a unified line of characters going down the left side of my terminal.
You can easily do it with sed:
ll | sed 's/./XXX&/'
Prepend XXXto all lines including empty lines.
ll | sed 's/^/XXX/'
Edit:
After changing your PS1 you can invoke the solution for all commands using
bash | sed 's/^/XXX/'
You could do something like this I suppose:
while read; do echo "xxx $REPLY"; done < <(ls -l)
Not really sure what the purpose of this would be though.
Using awk:
$ ll|awk '$0="XXX"$0'
There are multiple directories which contain a file with the same name:
direct_afaap/file.txt
direct_fgrdw/file.txt
direct_sardf/file.txt
...
Now I want to extract them to another directory, direct_new and with a different file name such as:
[mylinux~ ]$ ls direct_new/
file_1.txt file_2.txt file_3.txt
How can I do this?
BTW, if I want to put part of the name in original directory into the file name such as:
[mylinux~ ]$ ls direct_new/
file_afaap.txt file_fgrdw.txt file_sardf.txt
What can I do?
This little BaSH script will do it both ways:
#!/bin/sh
#
# counter
i=0
# put your new directory here
# can't be similar to dir_*, otherwise bash will
# expand it too
mkdir newdir
for file in `ls dir_*/*`; do
# gets only the name of the file, without directory
fname=`basename $file`
# gets just the file name, without extension
name=${fname%.*}
# gets just the extention
ext=${fname#*.}
# get the directory name
dir=`dirname $file`
# get the directory suffix
suffix=${dir#*_}
# rename the file using counter
fname_counter="${name}_$((i=$i+1)).$ext"
# rename the file using dir suffic
fname_suffix="${name}_$suffix.$ext"
# copy files using both methods, you pick yours
cp $file "newdir/$fname_counter"
cp $file "newdir/$fname_suffix"
done
And the output:
$ ls -R
cp.sh*
dir_asdf/
dir_ljklj/
dir_qwvas/
newdir/
out
./dir_asdf:
file.txt
./dir_ljklj:
file.txt
./dir_qwvas:
file.txt
./newdir:
file_1.txt
file_2.txt
file_3.txt
file_asdf.txt
file_ljklj.txt
file_qwvas.txt
while read -r line; do
suffix=$(sed 's/^.*_\(.*\)\/.*$/\1/' <<<$line)
newfile=$(sed 's/\.txt/$suffix\.txt/' <<<$line)
cp "$line" "~/direct_new/$newfile"
done <file_list.txt
where file_list is a list of your files.
You can achieve this with Bash parameter expansion:
dest_dir=direct_new
# dir based naming
for file in direct_*/file.txt; do
[[ -f "$file" ]] || continue # skip if not a regular file
dir="${file%/*}" # get the dir name from path
cp "$file" "$dest_dir/file_${dir#*direct_}.txt"
done
# count based naming
counter=0
for file in direct_*/file.txt; do
[[ -f "$file" ]] || continue # skip if not a regular file
cp "$file" "$dest_dir/file_$((++counter)).txt"
done
dir="${file%/*}" removes all characters starting from /, basically, giving us the dirname
${dir#*direct_} removes the direct_ prefix from dirname
((++counter)) uses Bash arithmetic expression to pre-increment the counter
See also:
Why you shouldn't parse the output of ls(1)
Get file directory path from file path
How to use double or single brackets, parentheses, curly braces
It may not be quite what you want, but it will do the job. Use cp --backup=numbered <source_file> <destination_directory:
$ find . -name test.sh
./ansible/test/integration/roles/test_command_shell/files/test.sh
./ansible/test/integration/roles/test_script/files/test.sh
./Documents/CGI/Code/ec-scripts/work/bin/test.sh
./Documents/CGI/Code/ec-scripts/trunk/bin/test.sh
./Test/test.sh
./bin/test.sh
./test.sh
$ mkdir BACKUPS
$ find . -name test.sh -exec cp --backup=numbered {} BACKUPS \;
cp: './BACKUPS/test.sh' and 'BACKUPS/test.sh' are the same file
$ ls -l BACKUPS
total 28
-rwxrwxr-x. 1 jack jack 121 Jun 9 10:29 test.sh
-rwxrwxr-x. 1 jack jack 34 Jun 9 10:29 test.sh.~1~
-rwxrwxr-x. 1 jack jack 34 Jun 9 10:29 test.sh.~2~
-rwxrwxr-x. 1 jack jack 388 Jun 9 10:29 test.sh.~3~
-rwxrwxr-x. 1 jack jack 388 Jun 9 10:29 test.sh.~4~
-rwxrwxr-x. 1 jack jack 20 Jun 9 10:29 test.sh.~5~
-rwxrwxr-x. 1 jack jack 157 Jun 9 10:29 test.sh.~6~
If you really want to put part of the folder name in, you have to decide exactly what part you want. You could, of course, just replace the directory separator character with some other character, and put the whole path into the filename.
how to do arithmetic operations with the size of the files and directory when they are in different unites like free space is in MB and the file size is in GB
With one preparatory command I am able to fetch the size of the "/home/abc/def" directory in MB. Its 30GB so getting in KB is not a good idea.
mount fssizeMB
===== =======
/home/abc/def 30002
root#abc:/home/abc/def> ls -lrth
total 7.0G
drwxrwxrwx 3 root root 114 Oct 29 2012 file1
drwxr-xr-x 3 root root 103 Nov 22 2012 file2
-rw-r--r-- 1 root root 114 Jan 25 2013 file3
-rw-r--r-- 1 mtc users 3.8G Jul 22 03:02 file4 <------------------- concerned file
-rw-r--r-- 1 mtc users 3.2G Jul 24 22:26 file5
-rw-r--r-- 1 root root 0 Jan 5 20:30 file6
How to turn below logic in script:
If twice the file size of file4 is < free space of "/home/abc/def " then echo success or else failure.
You could use stat or du -Sh to get the size of file (don't use ls for that in a script).
And to browse the files of a folder :
for i in <direcory>/*; do ...; done
Then, you could use test or [ commands (or [[ if you use Bash) to make a comparison (with -ge, -gt, -lt, -le options as arithmetic operators).
See the manpages of each command to get more information.
this would work with percentages, just to give you an idea, you could modify it to deal with MB or GB and so on.
my advice: doing arithmetic operations in bash is not such a good idea, you should work with programming languages that deal with special variable data type, like float or str and so on. bash is simpler and doesn't work so well with arithmetic operations. sure it does your + and -, but when it comes to percentages and floats... not so well.
try python or perl, or try researching something else. and definitely use, as suggested above, du -sh
#!/usr/bin/env bash
#take df -h and iterate trough percentages
#check to see if file system is full more than 80 %
perc="$(df -h | awk '{print $5}'| sed -e 's/[%/ a-z/A-Z].*//g' )"
#echo $perc
for p in $perc
do
if [ $p -gt 30 ] #change 30 to whatever
then
df -h | grep $p
echo -e "$p Exceeded on `hostname`"
fi
done
Most commands have options to show the size using a specific unit.
The -h flag of ls and df are to produce "human readable" format, which is not suitable for arithmetic calculations, as they can be in inconsistent units.
To get the size of a file, use stat, or even wc -c. It's a bad practice to parse the output of ls -l, so don't use that.
If you can get the size of a file consistently in kilobytes, and the size of free space consistently in bytes, not a problem, you can just multiply the size in bytes with 1024 to be able to make comparisons in consistent units.
The specific commands and flags to use will depend on your operating system and the software installed.
If I choose a zip file and right click "extract here" a folder with the zip filename is created and the entire content of the zip file is extracted into it.
However, I would like to convert several zip files via shell.
But when I do
unzip filename.zip
the folder "filename" is not created but all the files are extracted into the current directory.
I have looked at the parameters but there is no such parameter.
I also tried
for zipfile in \*.zip; do mkdir $zipfile; unzip $zipfile -d $zipfile/; done
but the .zip extension of the 2. $zipfile and 4. $zipfile have to be removed with sed.
If I do
for zipfile in \*.zip; do mkdir sed 's/\.zip//i' $zipfile; unzip $zipfile -d sed 's/\.zip//i' $zipfile/; done
it is not working.
How do I replace the .zip extension of $zipfile properly?
Is there an easier way than a shell script?
unzip file.zip -d xxx will extract files to directory xxx, and xxx will be created if it is not there. You can check the man page for details.
The awk line below should do the job:
ls *.zip|awk -F'.zip' '{print "unzip "$0" -d "$1}'|sh
See the test below,
note that I removed |sh at the end, since my zips are fake archives; I just want to show the generated command line here.
kent$ ls -l
total 0
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 001.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 002.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 003.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 004.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 005.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 006.zip
-rw-r--r-- 1 kent kent 0 Nov 12 23:10 007.zip
kent$ ls *.zip|awk -F'.zip' '{print "unzip "$0" -d "$1}'
unzip 001.zip -d 001
unzip 002.zip -d 002
unzip 003.zip -d 003
unzip 004.zip -d 004
unzip 005.zip -d 005
unzip 006.zip -d 006
unzip 007.zip -d 007
"extract here" is merely a feature of whatever unzip wrapper you are using. unzip will only extract what actually is in the archive. There is probably no simpler way than a shell script. But sed, awk etc. are not needed for this if you have a POSIX-compliant shell:
for f in *.zip; do unzip -d "${f%*.zip}" "$f"; done
(You MUST NOT escape the * or pathname expansion will not take place.) Be aware that if the ZIP archive contains a directory, such as with Eclipse archives (which always contain eclipse/), you would end up with ./eclipse*/eclipse/eclipse.ini in any case. Add echo before unzip for a dry run.
p7zip, the command line version of 7zip does the job
7z x '*.zip' -o'*'
On Debian and Ubuntu, you have to install p7zip-full.
Add the folder name and slash after the switch, example:
unzip -d newfolder/ zipfile.zip
zip will create the folder 'newfolder' and extract the archive into it.
Note that the trailing slash is not required but I guess it's just from old habit (I use Debian and Ubuntu).
aunpack from atool will do this by default for all sorts of archives.
In Termux bash I ended up with this, tested for spaces and dots in filenames. It only removes the last file extension and dot for the folder name. This assumes the zip file has a .zip extension otherwise it won't work. Went nuts trying to \ escape the quotes or use single quotes before realizing they just needed to be there as plain quote marks. Drop an echo in front of the zip command, or add a -l to audit the command. For some reason the position of the -d seems important for this implementation.
for f in *.zip; do unzip "$f" \-d "./${f%.*}/"; done; echo -end-
Open the terminal and locate the 00* files
cat filename.zip.00* > filename.zip
wait until the process ends, it depends on the file size.
The file joining process finished, now you can run the output file
unzip filename.zip