I'm trying to rename multiple files from the command line as follows:
for f in *.pdf
mv $f echo $f | sed 's/\( \)\([0-9]\)/-\2/'
end
I got the error:
mv: target 'filename' is not a directory
It is obvious that target's name has spaces and must be enclosed in quotes to be handled by mv command.
What should I do to get this script works?
Your script will try to move the file and the file called "echo" to the file, and pipe mv's output to sed.
What you want is to run that echo | sed in a command substitution, which fish denotes with ():
for f in *.pdf
mv $f (echo $f | sed 's/\( \)\([0-9]\)/-\2/')
end
It is obvious that target's name has spaces and must be enclosed in quotes to be handled by mv command.
It is not obvious because it doesn't need to be quoted. Fish does not perform word splitting like bash does. The filename is set once, and then $f will always yield the filename as one argument.
Quoting it like "$f" would be entirely superfluous.
For command substitutions, it splits them on newlines only, not spaces, so unless you have a filename with a newline (highly unlikely) you won't have a problem there either.
If you did you'd have to use string collect like
for f in *.pdf
mv $f (echo $f | sed 's/\( \)\([0-9]\)/-\2/' | string collect)
end
to ensure the command substitution results in one argument.
Related
I have a folder of 500 *.INI files that I need to manually edit. Within each INI file, I have the line Source =. I would like that line to become Source = C:\software\{filename}.
For instance, a dx4.ini file would need to be fixed to become: Source = C:\software\dx4
Is there a quick way to do this with Find, Grep, or Sed functions?
You can try with sed
For example
Input file contents:
file.txt
Source =
some lines..
script:
newstring='Source = C:\software\dx4'
oldstring='Source ='
echo `sed "s/$oldstring/$newstring/g" file.txt` > file.txt
After running the above commands
output:
Source = C:\software\dx4
some lines..
If you want to edit a file in a script, I think ed is the way to go. Combined with a shell for loop:
for file in *.INI; do
base=$(basename "$file" .INI)
ed -s "$file" <<EOF
/^Source =/s/=/= C:\\\\software\\\\$base/
w
EOF
done
(This does assume that filenames will not have newlines or ampersands in their names)
With GNU awk for the 3rd arg to match(), gensub(), and "inplace" editing:
awk -i inplace '
match($0,/(.*Source = C:\\software\\){filename}(.*)/,a) {
fname = gensub(/\..*/,"",1,FILENAME)
$0 = a[1] fname a[2]
}
1' *.INI
The above assumes you're running in a UNIX environment though your use of the term folder instead of directory and that path starting with C: and containing backslashes makes me suspicious. If you're on Windows then save the part between the 2 's (exclusive) in a file named foo.awk and execute it as awk -i inplace foo.awk *.INI or however it is you normally execute commands like this in Windows.
find *.ini -type -f > stack
while read line
do
sed -i s"#Source =#Source = C:\\software\\dx4#" "${line}"
done < stack
Assuming that a} You have sed with "-i" (the insert flag, which AFAIK is not always portable) and b} sed doesn't crap itself about a double escape sequence, I think that will work.
I have multiple files with lines like:
foo, 123456
bar, 654321
baz, 098765
I would like to remove everything on each line before (and including) the comma.
The output would be:
123456
654321
098765
I attempted to use the following after seeing something similar on another question, but the user didn't leave an explanation, so I'm not sure how the wildcard would be handled:
find . -name "*.csv" -type f | xargs sed -i -e '/*,/d'
Thank you for any help you can offer.
METHOD 1:
If it's always the 2nd column you want, you can do this with awk -- this command is actually splitting the rows on the whitespace rather than the comma, so it gets your second column -- the numbers, but without the leading space:
awk '{print $2}' < whatever.csv
METHOD 2:
Or to get everything after the comma (including the space):
sed -e 's/^.*,//g' < whatever.csv
METHOD 3:
If you want to find all of the .csv files and get the output of all of them together, you can do:
sed -e 's/^.*,//g' `find . -name '*.csv' -print`
METHOD 4:
Or the same way you were starting to -- with find and xargs:
find . -name '*.csv' -type f -print | xargs sed -e 's/^.*,//'
METHOD 5:
Making all of the .csv files into .txt files, processed in the way described above, you can make a brief shell script. Like this:
Create a script "bla.sh":
#!/bin/sh
for infile in `find . -name '*.csv' -print` ; do
outfile=`echo $infile | sed -e 's/.csv/.txt/'`
echo "$infile --> $outfile"
sed -e 's/^.*,//g' < $infile > $outfile
done
Make it executable by typing this:
chmod 755 bla.sh
Then run it:
./bla.sh
This will create a .txt output file with everything after the comma for each .csv input file.
ALTERNATE METHOD 5:
Or if you need them to be named .csv, the script could be updated like this -- this just makes an output file named "file-new.csv" for each input file named "file.csv":
#!/bin/sh
for infile in `find . -name '*.csv' -print` ; do
outfile=`echo $infile | sed -e 's/.csv/-new.csv/'`
echo "$infile --> $outfile"
sed -e 's/^.*,//g' < $infile > $outfile
done
Something like this should work for a single file. Let's say the
input is 'yourfile' and you want the output to go to 'outfile'.
sed 's/^.*,//' < yourfile > outfile
The syntax to do a search-and-replace is s/input_pattern/replacement/
The ^ anchors the input pattern to the beginning of the line.
A dot . matches any single character; .* matches a string of zero or more of any character.
The , matches the comma.
The replacement pattern is empty, so whatever matched the input_pattern
will be removed.
I am porting a sh script that was apparently written using GNU implementation of sed to BSD implementation of sed. The exact line in the script with the original comment are:
# escape dot in file extension to grep it
ext="$(echo $ext | sed 's/\./\\./' -)"
I am able to reproduce a results with the following (obviously I am not exhausting all possibilities values for ext) :
ext=.h; ext="$(echo $ext | sed 's/\./\\./' -)"; echo [$ext]
Using GNU's implementation of sed the following is returned:
[\.h]
Using BSD's implementation of sed the following is returned:
sed: -: No such file or directory
[]
Executing ext=.h; ext="$(echo $ext | sed 's/\./\\./')"; echo [$ext] returns [\.h] for both implementation of sed.
I have looked at both GNU and BSD's sed's man page have not found anything about the trailing "-". Googling for sed with a "-" is not very fruitful either.
Is the "-" a typo?
Is the "-" needed for some an unexpected value of $ext?
Is the issue not with sed, but rather with sh?
Can someone direct me to what I should be looking at, or even better, explain what the purpose of the "-" is?
On my system, that syntax isn't documented in the man page, but it is in the
'info' page:
sed OPTIONS... [SCRIPT] [INPUTFILE...]
If you do not specify INPUTFILE, or if INPUTFILE is -',sed'
filters the contents of the standard input.
Given that particular usage, I think you could leave off the '-' and it should
still work.
You got your specific question answered BUT your script is all wrong. Take a look at this:
# escape dot in file extension to grep it
ext="$(echo $ext | sed 's/\./\\./')"
The main problems with that are:
You're not quoting your variable ($ext) so it will go through file name expansion plus if it contains spaces will be passed to echo as multiple arguments instead of 1. Do this instead:
ext="$(echo "$ext" | sed 's/\./\\./')"
You're using an external command (sed) and a pipe to do something the shell can do trivially itself. Do this instead:
ext="${ext/./\.}"
Worst of all: You're escaping the RE meta-character (.) in your variable so you can pass it to grep to do an RE search on it as if it were a string - that doesn't make any sense and becomes intractable in the general case where your variable could contain any combination of RE metacharacters. Just do a string search instead of an RE search and you don't need to escape anything. Don't do either of the above substitution commands and then do either of these instead of grep "$ext" file:
grep -F "$ext" file
fgrep "$ext" file
awk -v ext="$ext" 'index($0,ext)' file
I was trying to do some quick filename cleanup at the shell (zsh, if it matters). Renaming files. (I'm using cp instead of mv just to be safe)
foreach f (\#*.ogg)
cp $f `echo $f | perl -pe 's/\#\d+ (.+)$/"\1"/'`
end
Now, I know there are tools to do stuff like this, but for personal interest I'm wondering how I can do it this way. Right now, I get an error:
cp: target `When.ogg"' is not a directory
Where 'When.ogg' is the last part of the filename. I've tried adding quotes (see above) and escaping the spaces, but nonetheless this is what I get.
Is there a reason I can't use the output of s perl pmr=;omrt as the final argument to another command line tool?
It looks like you have a space in the file names being processed, so each of your cp command lines evaluates to something like
cp \#nnnn When.Ogg When.ogg
When the cp command sees more than two arguments, the last one must be a target directory name for all the files to be copied to - hence the error message. Because your source filename ($f) contains a space it is being treated as two arguments - cp sees three args, rather than the two you intend.
If you put double quotes around the first $f that should prevent the two 'halves' of the name from being treated as separate file names:
cp "$f" `echo ...
This is what you need in bash, hope it's good for zsh too.
cp "$f" "`echo $f | perl -pe 's/\#\d+ (.+)$/\1/'`"
If the filename contains spaces, you also have quote the second argument of cp.
I often use
dir /b ... | perl -nle"$o=$_; s/.../.../; $n=$_; rename $o,$n if !-e $n"
The -l chomps the input.
The -e check is to avoid accidentally renaming all the files to one name. I've done that a couple of times.
In bash (and I'm guessing zsh), that would be
foreach f (...)
echo "$f" | perl -nle'$o=$_; s/.../.../; $n=$_; rename $o,$n if !-e $n'
end
or
find -name '...' -maxdepth 1 \
| perl -nle'$o=$_; s/.../.../; $n=$_; rename $o,$n if !-e $n'
or
find -name '...' -maxdepth 1 -exec \
perl -e'for (#ARGV) {
$o=$_; s/.../.../; $n=$_;
rename $o,$n if !-e $n;
}' {} +
The last supports file names with newlines in them.
use sed to replace every occurrence of /dir with $dir (replace / with $) in every script in a directory.
sed "s#/dir#$dir#g"
The $ keeps being interpreted as a function or variable call.
Is there a way around this?
thanks
Read your shell's friendly manual:
man sh
In the shell, "double quotes" around text allow variable interpretation inside, while 'single quotes' do not, a convention adopted by later languages such as Perl and PHP (but not e.g. JavaScript).
sed 's#/dir#$dir#g' *
To perform the replacement within the scripts do something like
find * -maxdepth 0 -type f | while read f; do mv $f $f.old && sed 's#/dir#$dir#' $f.old > $f; done
or just
perl -pi.old -e 's#/dir#\$dir#' * # Perl also interpolates variables in s commands
You can simply escape it with a backslash:
sed "s#/dir#\$dir#g"
shell approach
for file in file*
do
if [ -f "$file ];then
while read -r line
case "$line" in
*/dir* ) line=${line///dir/\$dir}
esac
echo $line > temp
done < "file"
mv temp $file
fi
done