Bash: how to traverse directory structure and execute commands?

流过昼夜 提交于 2019-12-01 03:53:19

For this kind of thing I always use find together with xargs:

$ find output-* -name "*.chunk.??" | xargs -I{} ./myexecutable -i {} -o {}.processed

Now since your script processes only one file at a time, using -exec (or -execdir) directly with find, as already suggested, is just as efficient, but I'm used to using xargs, as that's generally much more efficient when feeding a command operating on many arguments at once. Thus it's a very useful tool to keep in one's utility belt, so I thought it ought to be mentioned.

nikudesu

Something like:

for x in `find /home/brianonly -type f`
do
./yourexecutable -i $x -o $x.processed
done

As others have suggested, use find(1):

# Find all files named 'myfile.chunk.*' but NOT named 'myfile.chunk.*.processed'
# under the directory tree rooted at base-directory, and execute a command on
# them:
find base-directory -name 'output.*' '!' -name 'output.*.processed' -exec ./myexecutable -i '{}' -o '{}'.processed ';'

From the information provided, it sounds like this would be a completely straightforward translation of your C# idea.

for i in /home/brianly/output-*; do
    for j in "$i/"*.[0-9][0-9]; do
        ./myexecutable -i "$j" -o "$j.processed"
    done
done

That's what the find command is for.

http://linux.die.net/man/1/find

Use find and exec. Have a look at following

http://tldp.org/LDP/abs/html/moreadv.html

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