Parsing specific parts of output

Question

Basically, I have this script that updates wifi firmware on jailbroken devices (iOS). I need to check the output of this:

/usr/libexec/wifiFirmwareLoader -f

This outputs me the file path to the correct files needed for the rest of the script, but I don't want the whole path, I only want the name of the file at the end of the path. How could I go about parsing this?

If you need an example, this is what the output looks like.

RESOLVED /usr/share/firmware/wifi/C748383-A3/hans.trx

I would only want to store hans.trx in a variable.

But what did you try to solve this problem of yours? Post your research efforts into the question even if they are trivial — Inian, Dec 06 '17 at 12:12
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Veda · Answer 1 · 2017-12-06T20:09:23.807

You can use regexes in grep:

echo "RESOLVED /usr/share/firmware/wifi/C748383-A3/hans.trx" | grep -Eo "[^/]+$"

-E uses extended regexes. You could also use -P for perl regexes
-o only shows the matched output
[^/] matches everything that is not a /, the + makes it at least 1 character and the $ requires it to be at the end of the line
If this matches more lines that you would need, you could use postive lookahead (?=something) or positive lookbehind something\K to make the regex more strict (will need -P). Or in some cases I like to double grep (first grep for lines containing something, then get the part from the line you want)

Your final script could look something like this:

OUTPUT=$(/usr/libexec/wifiFirmwareLoader -f)
FILENAMES=$(echo "${OUTPUT}" | grep -Eo "[^/]+$")
echo "${FILENAMES}"

Parsing specific parts of output

1 Answers1