I have lines like these, and I want to know how many lines I actually have...
09:16:39 AM all 2.00 0.00 4.00 0.00 0.00 0.00 0.00 0.00 94.00
09:16:40 AM all 5.00 0.00 0.00 4.00 0.00 0.00 0.00 0.00 91.00
09:16:41 AM all 0.00 0.00 4.00 0.00 0.00 0.00 0.00 0.00 96.00
09:16:42 AM all 3.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 96.00
09:16:43 AM all 0.00 0.00 1.00 0.00 1.00 0.00 0.00 0.00 98.00
09:16:44 AM all 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 100.00
09:16:45 AM all 2.00 0.00 6.00 0.00 0.00 0.00 0.00 0.00 92.00
Is there a way to count them all using linux commands?
If you want to check the total line of all the files in a directory ,you can use find and wc:
I just made a program to do this ( with
node
)https://github.com/danschumann/gimme-lines/tree/master
grep -oE '\d+'
: In order to return the digit numbers ONLY.Use
wc
:This will output the number of lines in
<filename>
:Or, to omit the
<filename>
from the result usewc -l < <filename>
:You can also pipe data to
wc
as well:The tool
wc
is the "word counter" in UNIX and UNIX-like operating systems, you can also use it to count lines in a file, by adding the-l
option, sowc -l foo
will count the number of lines infoo
. You can also pipe output from a program like this:ls -l | wc -l
, which will tell you how many files are in the current directory.I've been using this:
I prefer it over the accepted answer because it does not print the filename, and you don't have to use
awk
to fix that. Accepted answer:wc -l myfile.txt
But I think the best one is GGB667's answer:
wc -l < myfile.txt
I will probably be using that from now on. It's slightly shorter than my way. I am putting up my old way of doing it in case anyone prefers it. The output is the same with those two methods.