How to get “wc -l” to print just the number of lines without file name?

问题

wc -l file.txt

outputs number of lines and file name.

I need just the number itself (not the file name).

I can do this

 wc -l file.txt | awk '{print $1}'

But maybe there is a better way?

回答1:

Try this way:

wc -l < file.txt

回答2:

cat file.txt | wc -l

According to the man page (for the BSD version, I don't have a GNU version to check):

If no files are specified, the standard input is used and no file name is displayed. The prompt will accept input until receiving EOF, or [^D] in most environments.

回答3:

To do this without the leading space, why not:

wc -l < file.txt | bc

回答4:

How about

wc -l file.txt | cut -d' ' -f1

i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)

回答5:

Comparison of Techniques

I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.

FOO="bar"
echo -n "$FOO" | wc -c                          # "       3"    (x)
echo -n "$FOO" | wc -c | bc                     # "3"           (√)
echo -n "$FOO" | wc -c | tr -d ' '              # "3"           (√)
echo -n "$FOO" | wc -c | awk '{print $1}'       # "3"           (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1         # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8         # "3"           (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//'    # "3"           (√)
echo -n "$FOO" | wc -c | grep -ch '^'           # "1"           (x)
echo $( printf '%s' "$FOO" | wc -c )            # "3"           (√)

I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.

bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.

Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).

回答6:

How about

grep -ch "^" file.txt

回答7:

Obviously, there are a lot of solutions to this. Here is another one though:

wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"

This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].

回答8:

Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)

below is the command :

find <directory path>  -type f | awk  'END{print NR}'

example : - find /tmp/ -type f | awk 'END{print NR}'

回答9:

This works for me using the normal wc -l and sed to strip any char what is not a number.

wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"

# 9249133

来源：https://stackoverflow.com/questions/10238363/how-to-get-wc-l-to-print-just-the-number-of-lines-without-file-name