How can I copy a variable to another variable in a shell script?
Assuming the user has passed in $1, how can its value be copied to another variable?
I assume it would look something like this...
cp $1 $2
echo "Copied value: $2"
5 Answers
Note that cp is used to copy files and directories. To define variables, you just have to use the following syntax:
v=$1
Example
$ cat a
echo "var v=$v"
v=$1
echo "var v=$v"
$ ./a 23 <---- we execute the script
var v= <---- the value is not set
var v=23 <---- the value is already set
Comments
Firstly cp is for copying files and directories only (as the man page states)
Secondly, it is not possible to assign to an argument variable ($0..$1..$n). They are meant to be read only.
You can do this instead:
input2=$1
It will copy the value of $1 to a new variable called $input2
Comments
i've found set -- to be a very useful command to set the positional parameters. e.g. in the example you gave, and well answered:
cp file1 file2
copies "file1" to "file2". frequently when i work with a few files, I'll do this instead:
set -- file1 file2
cp $1 $2
and if you want to reverse the names in the variables:
set -- $2 $1 # puts the current "$2" value in "$1", and vice versa, then
cp $1 $2 # copies what was file2 contents back to file1.
this without using any "named" variables, which you've already seen. my more common use is like:
set -- ${1%.txt} # strips a ".txt" suffix
set -- $1 $1.out $1.err # sets 2nd to <whatever>.out and 3rd to <whatever>.err, so
cmd $1.txt > $2 2>$3 # puts stdout in ...out and stderr in ...err
v
cp? what's wrong withx="$1"then echo "copied value:$x"cpthis year.