Some grep Help

[souled]

Alright, I need a little bit of help. How can I use grep to show only certain words in a line rather than the whole line? Or can it not be done with grep?

Here is what I'm trying to do.
If I use this command:
hellanzb.py status | grep %
I get this output:
492.0KB/s, 2736 MB queued, ETA: 01:34:56 (42%)

I want to be able to just extract the % done (without the % sign). In this case, I would want 42 (the answer to life?) to be displayed. What is the best way to do this?

Thanks for the help.

[fritz_monroe]

I'm definately not an expert, but how about using the -o switch? It's supposed to display only the portion of the line that matches the pattern you supply.

[souled]

Yeah, I just saw that switch, but it doesn't display everything. If I grep for %, then it will only reply with '%.' The number in front is what I need.

Also, Is there a way to knock off leading white space in a line through grep?

[Third Thoughts]

If you just want to display the number and the percent sign how about

grep -o [[:alnum:]]*%
I gave it a few test runs and it dealt with whitespace as well as random numbers not next to percent signs.

[souled]

Thanks. Exactly what I need.

[Arndt]

Also, Is there a way to knock off leading white space in a line through grep?

Can you be more specific? There probably is a way to do what you want, with 'sed' and 'perl' if not with 'grep', but I don't see exactly what you want.

[Third Thoughts]

Can you be more specific? There probably is a way to do what you want, with 'sed' and 'perl' if not with 'grep', but I don't see exactly what you want.

I think I know how you could knock off leading whitespace with grep. If you're only dealing with letters and numbers then you can type

grep -o [[:alnum:]].*
If you throw punctuation in its slightly more complicated, but not bad.

grep -o [[:alnum:][:punct:]].*
Its the same basic principle. Just set grep so it only displays the matching string and make it so that you match any string that doesn't start with a space.

~Andrew S.

[zentagonist]

Is there any way to do something similar without using the -o switch? And if not, is there anything other than grep that can do it?

If I have a file that contains, for example:


dog54 is still loose
dog76 is agressive
dog167 has been captured

and all I want is the list of dogs, ignoring everything else in the file ... so the output would be:



dog54
dog76
dog167

is there any way to do this, grepping or otherwise?

[ghostdog74]

is there any way to do this, grepping or otherwise?

Have you heard of awk?


awk '{
for( i=1;i<=NF;i++ )
{
if($i ~ /dog/) {
print $i;
}
}
}' "file"

output:


# ./test.sh
dog54
dog76
dog167

[geirha]

for the doglist I would use cut.

cut -d" " -f1 the_file.txt
with your example:

$ cat << EOF | cut -d" " -f1
> dog54 is still loose
> dog76 is agressive
> dog167 has been captured
> EOF
dog54
dog76
dog167

[ghostdog74]

for the doglist I would use cut.

cut -d" " -f1 the_file.txt
with your example:

$ cat << EOF | cut -d" " -f1
> dog54 is still loose
> dog76 is agressive
> dog167 has been captured
> EOF
dog54
dog76
dog167


yes, it does work, but only if all the dogs are in the same columns. It won't work if the dogs are elsewhere.

[geirha]

Some more information on how the lines may differ would help, though from this information, it sounds like grep is the best way to go:

egrep -o "dog[[:digit:]]+"

[ghostdog74]

Some more information on how the lines may differ would help, though from this information, it sounds like grep is the best way to go:

egrep -o "dog[[:digit:]]+"

how about this

egrep -o "(dog|dog[[:digit:]]+)" file

in case there is no numbers behind dog

[geirha]

[[:digit:]]+ means 1 or more digits
[[:digit:]]* means 0 or more digits
so in that case this is better
egrep -o "dog[[:digit:]]*"