How to count lines in a document?

Question

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?

Answer 1

Above are the preferred method but "cat" command can also helpful:

cat -n <filename>

Will show you whole content of file with line numbers.

Answer 2

Use wc:

wc -l <filename>

Answer 3

Or count all lines in subdirectories with a file name pattern (e.g. logfiles with timestamps in the file name):

wc -l ./**/*_SuccessLog.csv

Answer 4

there are many ways. using wc is one.

wc -l file

others include

awk 'END{print NR}' file

sed -n '$=' file (GNU sed)

grep -c ".*" file

Answer 5

wc -l <file.txt>

Or

command | wc -l

Answer 6

I know this is old but still: Count filtered lines

My file looks like:

Number of files sent
Company 1 file: foo.pdf OK
Company 1 file: foo.csv OK
Company 1 file: foo.msg OK
Company 2 file: foo.pdf OK
Company 2 file: foo.csv OK
Company 2 file: foo.msg Error
Company 3 file: foo.pdf OK
Company 3 file: foo.csv OK
Company 3 file: foo.msg Error
Company 4 file: foo.pdf OK
Company 4 file: foo.csv OK
Company 4 file: foo.msg Error

If I want to know how many files are sent OK:

grep "OK" <filename> | wc -l

OR

grep -c "OK" filename