问题
I'm using rsync to copy specific files from a source directory (and subdirectories) to a destination directory (and subdirectories). The mapping of the subdirectories is not identical, so I'm defining arrays of subdirectories of the destination directory that contain the source file paths.
I've been unable to successfully loop over the destination arrays with access to the names of the arrays. Here's a MWE (filenames and directories must be edited, obviously).
#!/bin/bash
sourcedir=~/Dropbox/230/
destdir=~/Dropbox/230/website/
# Destination arrays
organization=(
syllabus/syllabus.pdf
)
notes=(
notes/ME230_2014S_Part0_Lec00.pdf
notes/ME230_2014S_Part1_Lec01_partial.pdf
)
echo "synchronizing: rsyncing from $sourcedir to $destdir"
for destsubdir in $organization $notes
do
for sourcesubdir in "${destsubdir[@]}"
do
echo "from $sourcedir$sourcesubdir to $destdir$destsubdir/"
rsync -avz "$sourcedir$sourcesubdir" "$destdir$destsubdir/"
done
done
This is obviously wrong because I don't want the destination to be the contents of $destsubdir
but the name of the variable itself. I've been unable to successfully achieve this.
Note that a proper solution would allow any number of arrays to included without restructuring (e.g. no extra for-loops).
If multi-indexed arrays were possible in bash, this would be much easier, I think. Thanks for any help!
Solution
@jeanrjc has a good solution. Here's what it looks like in terms of my example above.
#!/bin/bash
sourcedir=~/Dropbox/230/
destdir=~/Dropbox/230/website/
# Destination arrays
organization=(
syllabus/syllabus.pdf
)
notes=(
notes/ME230_2014S_Part0_Lec00.pdf
notes/ME230_2014S_Part1_Lec01_partial.pdf
)
destsubdirArray=(
organization
notes
)
echo "synchronizing: rsyncing from $sourcedir to $destdir"
for destsubdir in ${destsubdirArray[*]}
do
destsubdirContents=$destsubdir[@];
for sourcesubdir in ${!destsubdirContents}
do
echo "from $sourcedir$sourcesubdir to $destdir$destsubdir/"
rsync -avz "$sourcedir$sourcesubdir" "$destdir$destsubdir/"
done
done
回答1:
I'm not sure I get what you meant, but maybe this can help you :
one=( "one" 5 6 7 8 )
two=( "two" 11 22 33 )
three=( "tree" 777 888 999 )
all=( one two three )
for i in ${all[*]}; do
ref=$i[@];
for j in ${!ref}; do
echo $j
done
done
one
5
6
7
8
two
11
22
33
tree
777
888
999
回答2:
Bash 4.3 supports namerefs -- the expansion of a nameref is the expansion of the nameref's contents rather than the nameref itself (it may help to think of namerefs as being similar to symlinks).
declare -n destsubdir
for destsubdir in organization notes ; do
for sourcesubdir in "${destsubdir[@]}" ; do
echo "$sourcedir$sourcesubdir $destdir$sourcesubdir"
# rsync -avz "$sourcedir$sourcesubdir" "$destdir$sourcesubdir"
done
done
unset -n destsubdir
For earlier bash versions the nearest equivalent is, as kisp states, to eval the entire inner loop. The quoting requirements can get out of hand really quickly.
for destsubdir in organization notes ; do
eval 'for sourcesubdir in "${'"$destsubdir"'[@]}" ; do
echo "$sourcedir$sourcesubdir" "$destdir$sourcesubdir"
# rsync -avz "$sourcedir$sourcesubdir" "$destdir$sourcesubdir"
done'
done
Note: those rsync commands have not been tested.
回答3:
Solution to pass the variable names instead of the variables :
You can pass the variable names itself to the first for, then evaluate them in the second if needed.
like
for var_name in variable1 variable2 ; do
eval "myvar=\$$var_name"
Anyway, the single for loop can be used with the elements only if you use ${array[@]} everywhere.
来源:https://stackoverflow.com/questions/22840498/bash-script-with-double-for-loop-over-multiple-arrays-with-access-to-array-names