how to find and execute a file in bash script?

Question

Hi I am trying to us the find command and -exec to -execute a file file when found but it doesn't seem to work. here is some code:

 ssh $i find $DIRECTORY -name "version.sh" -exec sh {} \;

I am getting the error no arguments for -exec

thanks!

If the file contains a valid shebang line, you don't need the sh. — tripleee
– tripleee, Commented Aug 29, 2012 at 20:16

Ignacio Vazquez-Abrams · Accepted Answer · 2012-08-29 20:11:52Z

5

Your side gets a semicolon, but the other side doesn't.

... \\\;

answered Aug 29, 2012 at 20:11

Ignacio Vazquez-Abrams

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3 Comments

tripleee Over a year ago

+1 To be more elaborate; your shell expands metacharacters before the command is executed. If you want the ssh command line to contain shell metacharacters, you need to quote or escape them.

Ignacio Vazquez-Abrams Over a year ago

The ssh command does see the semicolon, but the shell on the other side consumes it as a command separator.

tripleee Over a year ago

Yeah, it's the backslash which needs treatment, like you show.

Stephane Rouberol · Accepted Answer · 2012-08-30 07:42:13Z

0

A more readable version could be:

ssh $i "find $DIRECTORY -name version.sh | xargs -r sh"

answered Aug 30, 2012 at 7:42

Stephane Rouberol

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