CODEWITHSUNDEEP
bashsolariscut
user2773624
user2773624

Reputation: 107

Keep the delimiter in the output of cut

I have a script that has uses cut to pick some information out of a search that contains absolute paths. It looks something like:

PLACE=$(grep foo flatfile.txt | cut -d '/' -f 1-6)

The output looks like this:

machine1:/path/to/where/foo/is
machine2:/another/path/to/find/foo

I need it to look like this:

machine1:/path/to/where/foo/is/
machine2:/another/path/to/find/foo/

This needs to be printed to the console at the end of the script with echo "$PLACE" or something like that. The output will always be at least 2 lines, but usually more.

I tried about everything I can think of with echo, but it either shows no output at all or gives the output:

grep: '/' is a directory

I am running bash 3.00 on Solaris, if that helps any. I would really like to K.I.S.S. this by just having something tacked onto the end of the cut command, and not having to monkey with sed or awk. But, if that is the only way, so be it.

Upvotes: 5

Views: 5865

Answers (3)

Digital Trauma
Digital Trauma

Reputation: 16016

You can use one sed command, instead of multiple other commands:

$ PLACE=$(sed -n '/foo/s:\(\([^/]\+/\)\{6\}\)\(.*\)\+:\1:p' flatfile.txt)
$ echo "$PLACE"
machine1:/path/to/where/foo/is/
machine2:/another/path/to/find/foo/
$

... and then I'm reminded of why sed sometimes makes me shudder :-/

Upvotes: 0

iruvar
iruvar

Reputation: 23374

Try the following. Note that $PLACE is not quoted. This should allow word-splitting to happen and each word (which corresponds to one line of your output, assuming no embedded spaces) is then printed by printf with a following / and newline. Haven't been able to test this on Bash 3 on Solaris though

printf "%s/\n" $PLACE

Upvotes: 0

michael501
michael501

Reputation: 1482

try this :

PLACE=$(grep foo flatfile.txt | cut -d '/' -f 1-6 | xargs -I "%" echo %/)

Upvotes: 4

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