UNIX for Dummies Questions & Answers

Using the below code I want to find all .sff files and extract them. This works but it seems very cheap. Is there a safer more efficient way to go about this?

Code:

#!/bin/bash

G1=(/home/dirone)

find ${G1} -type f -name \*.sff | xargs python /usr/local/bin/sff_extract.py

What do you mean by "cheap"? It is standard shell code. Are you trying to bulletproof the code? There is no error checking that I can see.

Thanks I just thought it looked as though I could get errors if conditions changed.

If the goal is to treat every conceivable pathname verbatim ...

xargs treats spaces, single quotes, double quotes, and perhaps also underscores specially. If any of those characters occurred in a filename, your xargs solution would misbehave.

There are xargs options to work with logical lines, but those modes of operation sacrifice a great deal of efficiency for an imperfect workaround (quotes would no longer be a problem, but leading/trailing spaces could be).

Some find/xarg implementations support delimiting pathnames with a null byte.

While the following is slightly more efficient since piping isn't required, the primary advantage is robustness: any legal filename, no matter how bizarre, will be passed intact:

Code:

find "${G1}" -type f -name \*.sff -exec python /usr/local/bin/sff_extract.py {} +

Regards,
Alister