CODEWITHSUNDEEP
bashfindexec
cifz
cifz

Reputation: 1078

Find a directory and launch an executable in it

I've to find a directory named 'A' and then launch an executable named 'B' that's in it that takes a path as argument and have a line with an execl() : 

    execl("./C","C",path,(char*)0);
    perror("Exec failed");

where C is in 'A' and has suid bit set.

. I've thought of something like: 

    find -name A -execdir {}/B path \

However what I get is: 

    Exec failed: Permission denied

What's wrong ? Launching B from A give me no errors.

Sorry if it is a stupid question, I'm really new to bash script. Any help is appreciated, thanks a lot.

Upvotes: 2

Views: 90

Answers (1)

Gilles 'SO- stop being evil'
Gilles 'SO- stop being evil'

Reputation: 107759

When you run B through that find command, the current directory is the directory containing A (i.e. the parent directory of A), not A.

You'll get the right directory if you run find -path '*/A/B' -execdir {} \;.

This may or may not be the right way to solve your real-world problem. In this example, B serves no purpose, so it's hard to guess what the real-world problem is. Did you consider sudo?

Upvotes: 5

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