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 $(date)"
(though date +"Today is %c" is probably a better way to do that particular thing).
Related
There are 3 txt files called
1.txt 2.txt 3.txt
I want to batch copy with the name
1.txt.cp 2.txt.cp 3.txt.cp
using the wildcard *
I entered the command cp *.txt *.txt.cp
but it wasn't working...
cp : target *.txt.cp : is not a directory
what was the problem???
Use: for i in *.txt; do cp "$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.
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.
I was working my way through a primer on Shell (Bash) Scripting and had the following doubt :
Why does not the following command print the contents of cp's directory : which cp | ls -l
Does not piping by definition mean that we pass the output of one command to another i.e. redirect the output ?
Can someone help me out ? I am a newbie ..
The output of which is being piped to the standard input of ls. However, ls doesn't take anything on standard input. You want it (I presume) to be passed as a parameter. There are a couple of ways of doing that:
which cp | xargs ls -l
or
ls -l `which cp`
or
ls -l $(which cp)
In the first example the xargs command takes the standard output of the previous previous command and makes each line a parameter to the command whose name immediately follows xargs. So, for instance
find / | xargs ls -l
will do an ls -l on each file in the filesystem (there are some issues with this with peculiarly named files but that's beyond the scope of this answer).
The remaining two are broadly equivalent and use the shell to do this, expanding the output from which into the command line for cp.
It would be,
$ ls -l $(which cp)
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
OR
$ which cp | xargs ls -l
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
To pass the output of one command as parameter of another command, you need to use xargs along with the pipe symbol.
From man xargs
xargs - build and execute command lines from standard input.xargs reads items
from the standard input, delimited by blanks (which can be protected
with double or single quotes or a backslash) or newlines, and executes
the command (default is /bin/echo) one or more times with any initial-
arguments followed by items read from standard input. Blank lines on
the standard input are ignored.
my grep command looks like this
zgrep -B bb -A aa "pattern" *
I would lke to have output as:
file1:line1
file1:line2
file1:line3
file1:pattern
file1:line4
file1:line5
file1:line6
</blank line>
file2:line1
file2:line2
file2:line3
file2:pattern
file2:line4
file2:line5
file2:line6
The problem is that its hard to distinguish when lines corresponding to the first found result end and the lines corresponding to the second found result start.
Note that although man grep says that "--" is added between contiguous group of matches. It works only when multiple matches are found in the same file. but in my search (as above) I am searching multiple files.
also note that adding a new blank line after every bb+aa+1 line won't work because what if a file has less than bb lines before the pattern.
pipe grep output through
awk -F: '{if(f!=$1)print ""; f=$1; print $0;}'
Pipe | any output to:
sed G
Example:
ls | sed G
If you man sed you will see
G Append's a newline character followed by the contents of the hold space to the pattern space.
The problem is that its hard to distinguish when lines corresponding to the first found result end and the lines corresponding to the second found result start.
Note that although man grep says that "--" is added between contiguous group of matches. It works only when multiple matches are found in the same file. but in my search (as above) I am searching multiple files.
If you don't mind a -- in lieu of a </blank line>, add the -0 parameter to your grep/zgrep command. This should allow for the -- to appear even when searching multiple files. You can still use the -A and -B flags as desired.
You can also use the --group-separator parameter, with an empty value, so it'd just add a new-line.
some-stream | grep --group-separator=
I can't test it with the -A and -B parameters so I can't say for sure but you could try using sed G as mentioned here on Unix StackEx. You'll loose coloring though if that's important.
There is no option for this in grep and I don't think there is a way to do it with xargs or tr (I tried), but here is a for loop that will do it (for f in *; do grep -H "1" $f && echo; done):
[ 11:58 jon#hozbox.com ~/test ]$ for f in *; do grep -H "1" $f && echo; done
a:1
b:1
c:1
d:1
[ 11:58 jon#hozbox.com ~/test ]$ ll
-rw-r--r-- 1 jon people 2B Nov 25 11:58 a
-rw-r--r-- 1 jon people 2B Nov 25 11:58 b
-rw-r--r-- 1 jon people 2B Nov 25 11:58 c
-rw-r--r-- 1 jon people 2B Nov 25 11:58 d
The -H is to display file names for grep matches. Change the * to your own file glob/path expansion string if necessary.
Try with -c 2; with printing a context I see grep is separating its found o/p
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