I'm trying to get the count of a matching pattern from a variable to check the count of it, but it's only returning 1 as the results, here is what I'm trying to do:
x="HELLO|THIS|IS|TEST"
echo $x | grep -c "|"
Expected result: 3
Actual Result: 1
Do you know why is returning 1 instead of 3?
Thanks.
grep -c counts lines not matches within a line.
You can use awk to get a count:
x="HELLO|THIS|IS|TEST"
echo "$x" | awk -F '|' '{print NF-1}'
3
Alternatively you can use tr and wc:
echo "$x" | tr -dc '|' | wc -c
3
$ echo "$x" | grep -o '|' | grep -c .
3
grep -c does not count the number of matches. It counts the number of lines that match. By using grep -o, we put the matches on separate lines.
This approach works just as well with multiple lines:
$ cat file
hello|this|is
a|test
$ grep -o '|' file | grep -c .
3
The grep manual says:
grep, egrep, fgrep - print lines matching a pattern
and for the -c flag:
instead print a count of matching lines for each input file
and there is just one line that match
You don't need grep for this.
pipe_only=${x//[^|]} # remove everything except | from the value of x
echo "${#pipe_only}" # output the length of pipe_only
Try this :
$ x="HELLO|THIS|IS|TEST"; echo -n "$x" | sed 's/[^|]//g' | wc -c
3
With only one pipe with perl:
echo "$x" |
perl -lne 'print scalar(() = /\|/g)'
Related
I was trying to filter all the files from the URLs and get only paths.
echo -e "http://sub.domain.tld/secured/database_connect.php\nhttp://sub.domain.tld/section/files/image.jpg\nhttp://sub.domain.tld/.git/audio-files/top-secret/audio.mp3" | grep -Ei "(http|https)://[^/\"]+" | sort -u
http://sub.domain.tld
But I want the result like this
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
Is there any way to do it with sed or grep
Using grep
$ echo ... | grep -o '.*/'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
with grep
If your grep has the -o option:
... | grep -Eio 'https?://.*/'
If there could be multiple URLs per line:
... | grep -Eio 'https?://[^[:space:]]+/'
with sed
If the input is always precisely one URL per line and nothing else, you can just delete the filename part:
... | sed 's/[^/]*$//'
You could use match function of awk, will work in any version of awk. Simple explanation would be, passing echo command's output to awk program. Using match matching everything till last occurrence of / and then printing the sub-string to print just before /(with -1 to RLENGTH).
your_echo_command | awk 'match($0,/.*\//){print substr($0,RSTART,RLENGTH-1)}'
GNU Awk
$ echo ... | awk 'match($0,/.*\//,a){print a[0]}'
$ echo ... | awk '{print gensub(/(.*\/).*/,"\\1",1)}'
$ echo ... | awk 'sub(/[^/]*$/,"")'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
xargs
$ echo ... | xargs -i sh -c 'echo $(dirname "{}")/'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
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 $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$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).
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)
How about
grep -ch "^" file.txt
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:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
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}'
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
This question already has answers here:
Get just the integer from wc in bash
(19 answers)
Closed 8 years ago.
I want to get only the number of lines in a file:
so I do:
$wc -l countlines.py
9 countlines.py
I do not want the filename, so I tried
$wc -l countlines.py | cut -d ' ' -f1
but this just echo empty line.
I just want number 9 to be printed
Use stdin and you won't have issue with wc printing filename
wc -l < countlines.py
You can also use awk to count lines. (reference)
awk 'END { print NR }' countlines.py
where countlines.py is the file you want to count
If your file doesn't ends with a \n (new line) the wc -l gives a wrong result. Try it with the next simulated example:
echo "line1" > testfile #correct line with a \n at the end
echo -n "line2" >> testfile #added another line - but without the \n
the
$ wc -l < testfile
1
returns 1. (The wc counts the number of newlines (\n) in a file.)
Therefore, for counting lines (and not the \n characters) in a file, you should to use
grep -c '' testfile
e.g. find empty character in a file (this is true for every line) and count the occurences -c. For the above testfile it returns the correct 2.
Additionally, if you want count the non-empty lines, you can do it with
grep -c '.' file
Don't trust wc :)
Ps: one of the strangest use of wc is
grep 'pattern' file | wc -l
instead of
grep -c 'pattern' file
cut is being confused by the leading whitespace.
I'd use awk to print the 1st field here:
% wc -l countlines.py | awk '{ print $1 }'
As an alternative, wc won't print the file name if it is being piped input from stdin
$ cat countlines.py | wc -l
9
yet another way :
cnt=$(wc -l < countlines.py )
echo "total is $cnt "
Piping the file name into wc removes it from the output, then translate away the whitespace:
wc -l <countlines.py |tr -d ' '
Use awk like this:
wc -l countlines.py | awk {'print $1'}
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 $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$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).
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)
How about
grep -ch "^" file.txt
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:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
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}'
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
Suppose I have the string 1:2:3:4:5 and I want to get its last field (5 in this case). How do I do that using Bash? I tried cut, but I don't know how to specify the last field with -f.
You can use string operators:
$ foo=1:2:3:4:5
$ echo ${foo##*:}
5
This trims everything from the front until a ':', greedily.
${foo <-- from variable foo
## <-- greedy front trim
* <-- matches anything
: <-- until the last ':'
}
Another way is to reverse before and after cut:
$ echo ab:cd:ef | rev | cut -d: -f1 | rev
ef
This makes it very easy to get the last but one field, or any range of fields numbered from the end.
It's difficult to get the last field using cut, but here are some solutions in awk and perl
echo 1:2:3:4:5 | awk -F: '{print $NF}'
echo 1:2:3:4:5 | perl -F: -wane 'print $F[-1]'
Assuming fairly simple usage (no escaping of the delimiter, for example), you can use grep:
$ echo "1:2:3:4:5" | grep -oE "[^:]+$"
5
Breakdown - find all the characters not the delimiter ([^:]) at the end of the line ($). -o only prints the matching part.
You could try something like this if you want to use cut:
echo "1:2:3:4:5" | cut -d ":" -f5
You can also use grep try like this :
echo " 1:2:3:4:5" | grep -o '[^:]*$'
One way:
var1="1:2:3:4:5"
var2=${var1##*:}
Another, using an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
var2=${var2[#]: -1}
Yet another with an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
count=${#var2[#]}
var2=${var2[$count-1]}
Using Bash (version >= 3.2) regular expressions:
var1="1:2:3:4:5"
[[ $var1 =~ :([^:]*)$ ]]
var2=${BASH_REMATCH[1]}
$ echo "a b c d e" | tr ' ' '\n' | tail -1
e
Simply translate the delimiter into a newline and choose the last entry with tail -1.
Using sed:
$ echo '1:2:3:4:5' | sed 's/.*://' # => 5
$ echo '' | sed 's/.*://' # => (empty)
$ echo ':' | sed 's/.*://' # => (empty)
$ echo ':b' | sed 's/.*://' # => b
$ echo '::c' | sed 's/.*://' # => c
$ echo 'a' | sed 's/.*://' # => a
$ echo 'a:' | sed 's/.*://' # => (empty)
$ echo 'a:b' | sed 's/.*://' # => b
$ echo 'a::c' | sed 's/.*://' # => c
There are many good answers here, but still I want to share this one using basename :
basename $(echo "a:b:c:d:e" | tr ':' '/')
However it will fail if there are already some '/' in your string.
If slash / is your delimiter then you just have to (and should) use basename.
It's not the best answer but it just shows how you can be creative using bash commands.
If your last field is a single character, you could do this:
a="1:2:3:4:5"
echo ${a: -1}
echo ${a:(-1)}
Check string manipulation in bash.
Using Bash.
$ var1="1:2:3:4:0"
$ IFS=":"
$ set -- $var1
$ eval echo \$${#}
0
echo "a:b:c:d:e"|xargs -d : -n1|tail -1
First use xargs split it using ":",-n1 means every line only have one part.Then,pring the last part.
Regex matching in sed is greedy (always goes to the last occurrence), which you can use to your advantage here:
$ foo=1:2:3:4:5
$ echo ${foo} | sed "s/.*://"
5
A solution using the read builtin:
IFS=':' read -a fields <<< "1:2:3:4:5"
echo "${fields[4]}"
Or, to make it more generic:
echo "${fields[-1]}" # prints the last item
for x in `echo $str | tr ";" "\n"`; do echo $x; done
improving from #mateusz-piotrowski and #user3133260 answer,
echo "a:b:c:d::e:: ::" | tr ':' ' ' | xargs | tr ' ' '\n' | tail -1
first, tr ':' ' ' -> replace ':' with whitespace
then, trim with xargs
after that, tr ' ' '\n' -> replace remained whitespace to newline
lastly, tail -1 -> get the last string
For those that comfortable with Python, https://github.com/Russell91/pythonpy is a nice choice to solve this problem.
$ echo "a:b:c:d:e" | py -x 'x.split(":")[-1]'
From the pythonpy help: -x treat each row of stdin as x.
With that tool, it is easy to write python code that gets applied to the input.
Edit (Dec 2020):
Pythonpy is no longer online.
Here is an alternative:
$ echo "a:b:c:d:e" | python -c 'import sys; sys.stdout.write(sys.stdin.read().split(":")[-1])'
it contains more boilerplate code (i.e. sys.stdout.read/write) but requires only std libraries from python.