I am trying to delete some files with loop. But can't figure out the way of doing it. Here's what I am trying to do:
#!/bin/bash dir1=/path/to/file dir2=/path/to/file dir3=/path/to/file for i in 1 2 3 do rm ${dir$i} done 2 Answers
This is possible with eval, but I'd strongly advice not to use that!
Use an array instead:
dirs=( "/path/to/file1" "/path/to/file2" "/path/to/file3" ) for i in 0 1 2 do rm "${dirs[$i]}" done # OR simply loop all the array values: for dir in "${dirs[@]}" do rm "$dir" done Note, that arrays are 0-based.
5You can do it in bash with indirection
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new pa‐ rameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion.
but it requires an intermediate variable ex. given:
dir1=/path/to/file1 dir2=/path/to/file2 dir3=/path/to/file3 then
$ for i in 1 2 3; do d=dir$i; echo rm "${!d}"; done rm /path/to/file1 rm /path/to/file2 rm /path/to/file3 You could eliminate the intermediate variable by looping over strings dir1 dir2 dir3 using brace expansion:
for d in dir{1..3}; do echo rm "${!d}"; done The same feature is available in zsh using the P modifier:
% for i in 1 2 3; do d=dir$i; echo rm ${(P)d}; done See Use a variable reference "inside" another variable
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