Is there any performance difference in the below shell commands:
find . -type f -empty -exec rm '{}' \;
find . -type f -empty -exec sh -c "/bin/rm {}" \;
Your 2nd command is going to be slower since it will spawn a sub shell for each entry found by find command.
However at the same time 2nd command will be more flexible in nature if you want to do some variable assignment etc like this:
find . -type f -empty -exec sh -c "x=1; /bin/rm {}" \;
Related
I wrote a simple command to find folders older than a month and delete them.
Here is the command :
find . -type d -mtime +31 -exec bash -c 'rm -rfv "$0"' {} \;
It works fine in most cases, but sometime, the -exec "ignores" the result.
After running the command once, If I run the find without -exec, it still finds some folders older than a month which have not been removed.
I then tried with a simple echo and got no output :
$ find . -type d -mtime +31
./folder_A
./Folder_B
$ find . -type d -mtime +31 -exec bash -c 'echo "$0"' {} \;
<No output>
I found a workround using grep but I'm wondering why the -exec ignores some results.
Anyone knows ?
Here is the workaround :
find . -type d -mtime +31 | grep . --color=never | while read line ; do rm -rvf "$line" ; done
Your find commands seem a bit strange to me. I'd suggest changing:
find . -type d -mtime +31 -exec bash -c 'rm -rfv "$0"' {} \;
to:
find . -type d -mtime +31 -exec rm -rfv {} \;
or even better (increases performance):
find . -type d -mtime +31 -exec rm -rfv {} +
If you insist on calling Bash explicitly (although I can't think of any reason why you should as rm is not a Bash built-in), I'd suggest changing:
find . -type d -mtime +31 -exec bash -c 'rm -rfv "$0"' {} \;
to:
find . -type d -mtime +31 -exec bash -c 'rm -rfv {}' \;
Same goes for the find command containing echo.
Any idea why this command is not working? btw, I'm trying to strip out the extensions of all csv files in current directory.
find -type f -iname "*.csv" -exec mv {} $(basename {} ".csv") \;
Tried many variants including the parameter expansions, xargs ... Even then all went futile.
This should do:
find ./ -type f -iname "*.csv" -exec sh -c 'mv {} $(basename {} .csv)' \;
find is able to substitute {} with its findings since the quotes prevent executing the subshell until find is done. Then it executes the -exec part.
The problem why yours is not working is that $(basename {} ".csv") is executed in a subshell (-> $()) and evaluated beforehand. If we look at the command execution step-by-step you will see what happens:
find -type f -iname "*.csv" -exec mv {} $(basename {} ".csv") \; - your command
find -type f -iname "*.csv" -exec mv {} {} \; - subshell gets evaluated ($(basename {} ".csv") returns {} since it interprets {} as a literal)
find -type f -iname "*.csv" -exec mv {} {} \; - as you see now: move does actually nothing
First, take care that you have no subdirectories; find, without extra arguments, will automatically recur into any directory below.
Simple approach: if you have a small enough number of files, just use the glob (*) operator, and take advantage of rename:
$ rename 's/.csv$//' *.csv
If you have too many files, use find, and perhaps xargs:
$ find . -maxdepth 1 -type f -name "*.csv" | xargs rename 's/.csv$//'
If you want to be really safe, tell find and xargs to delimit with null-bytes, so that you don't have weird filenames (e.g., with spaces or newlines) mess up the process:
$ find . -maxdepth 1 -type f -name "*.csv" -print0 | xargs -0 rename 's/.csv$//'
i have a piece of code that should print all files in specific dir. i use find exec for this:
find ${_di} -type f -print -exec Log "$(stat -c%y {}) - {}" \;
Where log is function of mine defines in same file.
But id does not work and i get error message:
"find: Log: No such file or directory".
Why? What is wrong in this piece of code?
Function can't be used in -exec however bash -c can be used as command.
Slightly modified to using + as -exec command terminator and {} last to allow to reduce the number of bash processes spawned.
find ${_di} -type f -print -exec bash -c "$(typeset -f Log)"$'\n''for arg; do Log "$(stat -c%y "$arg") - $arg"; done' -- {} +
the argument -- can be replaced by anything else it is used for $0 argument of shell.
bash -c 'echo $0' hello
Maybe -printf "%TY-%Tm-%Td %TT - %p\n" option could achieve the same result, more efficiently without launching other process.
Also using echo may be less safe than using find -print option, considering the following use case.
touch file.$'\e#8'
find . -type d ! -name . -prune -o -name file'*' -print
find . -type d ! -name . -prune -o -name file'*' -exec echo {} \;
You need to export the function, and then, as Nahuel says, run bash in the -exec:
$ export -f Log
$ find ${_di} -type f -exec bash -c 'Log "$(stat -c%y {}) - {}"' bash \;
I understand that using + rather than \; in a find command with -exec can speed things up because with \; the target of -exec is executed once for each result of the find command, whereas with + the target of -exec is executed "as needed."
This code works as expected and processes all subdirectories:
find "${directory}" -iname "*.jpg" -type d -prune -exec bash -c 'myscript "{}"' \;
But this code does NOT work:
find "${directory}" -iname "*.jpg" -type d -prune -exec bash -c 'myscript "$#"' bash {} +
It processes only one directory rather than all of them.
I'm obviously missing something about the proper syntax of using + when calling a function.
I want to run:
./my_script.py a_file &
... on all files in the current folder that end with .my_format, so I do:
find . -type f -name "*.my_format" -exec ./my_script {} & \;
but it doesn't work. How should I include & in the -exec parameter?
Try this:
$ find . -type f -name "*.my_format" -exec sh -c './my_script {} &' \;
The mostly likely reason your attempt didn't work is because find executes the command using one of the exec(3) family of standard c library calls which don't understand job control - the & symbol. The shell does understand the "run this command in the background" hence the -exec sh ... invocation
Try this find command:
find . -type f -name "*.my_format" -exec bash -c './my_script "$1" &' - '{}' \;