GNU find and more complex -exec option parameter values

Question

This works for me:

$> find . -name "*.log" -exec basename '{}' \;
20160114.log
20160115.log
20160116.log
20160117.log
20160118.log

Is the {}, \ and ';' mandatory when using -exec as any other syntax simply doesn't work?
The following, more complex example doesn't work:
```
$> find . -name "*.log" -exec echo $(basename '{}') \;
./log/20160114.log
./log/20160115.log
./log/20160116.log
./log/20160117.log
./log/20160118.log
```
echo here is just to demonstrate. I eventually plan to use something like rm $TARGET_DIR/$(basename '{}') in its place… It just doesn't work that way (nesting). Any ideas?

Regarding your first question, yes. The `{}` is replaced by the file name by `find`, the backslash is to escape the semi-colon which have a special meaning in shells, and the semi-colon tells `find` that the command has ended. — Some programmer dude, Jan 19 '16 at 07:36
How about using `xargs`, e.g. `find . -name "*.log" | xargs basename` — mnille, Jan 19 '16 at 07:39
@mnile, that works. I'm wondering here if there's a limitation of what can be put when using `-exec` format — msciwoj, Jan 19 '16 at 08:11
@msciwoj The issue isn't the nesting, it's the order of operations. `$()` is executed first. And `basename {}` (without `find` to replace the `{}` first), probably just produced `{}`, so it would execute `echo {}` for each file. You can however have it execute in a subshell, by simply doing `find . -name "*.log" -exec bash -c 'echo "$(basename "$1")"' bash {} \;` Or choose some other construct — Reinstate Monica Please, Jan 20 '16 at 02:55

0 Answers0