Solved Understanding the command find with the -exec parameter.

[emmex]

Hi,
the command:
# find /tmp -exec /root/bin/test.sh {} \;
with the /root/bin/test.sh script:

#!/bin/sh
#
echo "$1"
exit 1

Should stop at the first file found.
From the find() page:
-exec utility [argument ...] ;
True if the program named utility returns a zero value as its
exit status.

Why the command doesn't stop ?
--
Regards
Maurizio

 

[covacat]

won't stop just will eval to false (won't print or whatever)
like any other condition (bad size, bad name, bad timestamp)

 

[Kai Burghardt]

[…] True if the program named utility returns a zero value as its exit status. […]

This refers to the value the primary evaluates to. For example​

find /tmp -exec /root/bin/test.sh '{}' ';' -and -delete

will not delete any file because your script returns a non‑zero exit status. What you want is, taking account of short‑circuit evaluation,​

find /tmp -exec /root/bin/test.sh '{}' ';' -or -quit

 

[emmex]

Thank you Kai Burghardt, it works.