Linux commands to copy one file to many files - linux

Question

Is there a one-line command/script to copy one file to many files on Linux?
cp file1 file2 file3
copies the first two files into the third. Is there a way to copy the first file into the rest?

Answer 1

Does
cp file1 file2 ; cp file1 file3
count as a "one-line command/script"? How about
for file in file2 file3 ; do cp file1 "$file" ; done
?
Or, for a slightly looser sense of "copy":
tee <file1 file2 file3 >/dev/null

Answer 2

just for fun, if you need a big list of files:
tee <sourcefile.jpg targetfiles{01-50}.jpg >/dev/null- Kelvin Feb 12 at 19:52
But there's a little typo. Should be:
tee <sourcefile.jpg targetfiles{01..50}.jpg >/dev/null
And as mentioned above, that doesn't copy permissions.

Answer 3

You can improve/simplify the for approach (answered by #ruakh) of copying by using ranges from bash brace expansion:
for f in file{1..10}; do cp file $f; done
This copies file into file1, file2, ..., file10.
Resource to check:
http://wiki.bash-hackers.org/syntax/expansion/brace#ranges

Answer 4

for FILE in "file2" "file3"; do cp file1 $FILE; done

Answer 5

You can use shift:
file=$1
shift
for dest in "$#" ; do
cp -r $file $dest
done

Answer 6

cat file1 | tee file2 | tee file3 | tee file4 | tee file5 >/dev/null

Answer 7

(no loops used)
To copy the content of one file (fileA.txt) to many files (fileB.txt, fileC.txt, fileD.txt) in Linux,
use the following combination cat and tee commands:
cat fileA.txt | tee fileB.txt fileC.txt fileD.txt >/dev/null
applicable to any file extensions
only file names and extensions change, everything else remains same.

Answer 8

Use something like the following. It works on zsh.
cat file > firstCopy > secondCopy > thirdCopy
or
cat file > {1..100} - for filenames with numbers.
It's good for small files.
You should use the cp script mentioned earlier for larger files.

Answer 9

I'd recommend creating a general use script and a function (empty-files), based on the script, to empty any number of target files.
Name the script copy-from-one-to-many and put it in your PATH.
#!/bin/bash -e
# _ _____
# | |___ /_ __
# | | |_ \ \/ / Lex Sheehan (l3x)
# | |___) > < https://github.com/l3x
# |_|____/_/\_\
#
# Copy the contents of one file to many other files.
source=$1
shift
for dest in "$#"; do
cp $source $dest
done
exit
NOTES
The shift above removes the first element (the source file path) from the list of arguments ("$#").
Examples of how to empty many files:
Create file1, file2, file3, file4 and file5 with content:
for f in file{1..5}; do echo $f > "$f"; done
Empty many files:
copy-from-one-to-many /dev/null file1 file2 file3 file4 file5
Empty many files easier:
# Create files with content again
for f in file{1..5}; do echo $f > "$f"; done
copy-from-one-to-many /dev/null file{1..5}
Create empty_files function based on copy-from-one-to-many
function empty-files()
{
copy-from-one-to-many /dev/null "$#"
}
Example usage
# Create files with content again
for f in file{1..5}; do echo $f > "$f"; done
# Show contents of one of the files
echo -e "file3:\n $(cat file3)"
empty_files file{1..5}
# Show that the selected file no longer has contents
echo -e "file3:\n $(cat file3)"
Don't just steal code. Improve it; Document it with examples and share it. - l3x
Here's a version that will preface each cp command with sudo:
#!/bin/bash -e
# Filename: copy-from-one-to-may
# _ _____
# | |___ /_ __
# | | |_ \ \/ / Lex Sheehan (l3x)
# | |___) > < https://github.com/l3x
# |_|____/_/\_\
#
# Copy the contents of one file to many other files.
# Pass --sudo if you want each cp to be perfomed with sudo
# Ex: copy-from-one-to-many $(mktemp) /tmp/a /tmp/b /tmp/c --sudo
if [[ "$*" == *--sudo* ]]; then
maybe_use_sudo=sudo
fi
source=$1
shift
for dest in "$#"; do
if [ $dest != '--sudo' ]; then
$maybe_use_sudo cp $source $dest
fi
done
exit

Answer 10

You can use standard scripting commands for that instead:
Bash:
for i in file2 file3 ; do cp file1 $i ; done

Answer 11

The simplest/quickest solution I can think of is a for loop:
for target in file2 file3 do; cp file1 "$target"; done
A dirty hack would be the following (I strongly advise against it, and only works in bash anyway):
eval 'cp file1 '{file2,file3}';'

Answer 12

Go with the fastest cp operations
seq 1 10 | xargs -P 0 -I xxx cp file file-xxx
it means
seq 1 10 count from 1 to 10
| pipe it xargs
-P 0 do it in parallel - as many as needed
-I xxx name of each input xargs receives
cp file file-xxx means copy file to file-1, file-2, etc
and if name of files are different here is the other solutions.
First have the list of files which are going to be created. e.g.
one
two
three
four
five
Second save this list on disk and read the list with xargs just like before but without using seq.
xargs -P 0 -I xxx cp file xxx < list
which means 5 copy operations in parallel:
cp file one
cp file two
cp file three
cp file four
cp file five
and for xargs here is the behind the scene (5 forks)
3833 pts/0 Ss 0:00 bash
15954 pts/0 0:00 \_ xargs -P 0 -I xxx cp file xxx < list
15955 pts/0 0:00 \_ cp file one
15956 pts/0 0:00 \_ cp file two
15957 pts/0 0:00 \_ cp file three
15958 pts/0 0:00 \_ cp file four
15959 pts/0 0:00 \_ cp file five

Answer 13

I don't know how correct this is but i have used something like this
echo ./file1.txt ./file2.txt ./file3.txt | xargs -n 1 cp file.txt
Where echo ./file1.txt ... is destination of a file and use it to feed xargs with one "destination" by one. Therefore command xargs -n 1. And lastly cp file.txt, which is self explanatory i think :)