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bash echo number of lines of file given in a bash variable
Was wondering how you output the number of lines in a text file to screen and then store it in a variable.
I have a file called stats.txt and when I run wc -l stats.txt it outputs 8 stats.txt
I tried doing x = wc -l stats.txt thinking it would store the number only and the rest is just for visual but it does not work :(
Thanks for the help
There are two POSIX standard syntax for doing this:
x=`cat stats.txt | wc -l`
or
x=$(cat stats.txt | wc -l)
They both run the program and replace the invocation in the script with the standard output of the command, in this case assigning it to the $x variable. However, be aware that both trim ending newlines (this is actually what you want here, but can be dangerous sometimes, when you expect a newline).
Also, the second case can be easily nested (example: $(cat $(ls | head -n 1) | wc -l)). You can also do it with the first case, but it is more complex:
`cat \`ls | head -n 1\` | wc -l`
There are also quotation issues. You can include these expressions inside double-quotes, but with the back-ticks, you must continue quoting inside the command, while using the parenthesis allows you to "start a new quoting" group:
"`echo \"My string\"`"
"$(echo "My string")"
Hope this helps =)
you may try:
x=`cat stats.txt | wc -l`
or (from the another.anon.coward's comment):
x=`wc -l < stats.txt`
Related
I am interested in counting the number of lines in the output of ps command which I can do with
echo "$(ps | wc -l)"
However, I now have the output of ps command stored in a variable by doing X="$(ps)". How do I pass X to wc -l command without using a pipe? I tried $(wc -l < "$X") but it didn't work. I read the man page for wc and it takes a file as an argument. So I guess another way to frame the question would be - How do I treat value of a variable as a file to pass as an argument to a command in bash script?
I am fairly new to bash scripting and keywords I tried to search with didn't give clear answer to my questions.
I suggest:
echo "$X" | wc -l
or
wc -l <<< "$X"
I am new to bash and scripting and I am trying to create a simple script but for some reason it won't let me run this:
fileCount= ls -1 | wc -l
#echo $fileCount
for (( i=0; i<$fileCount; ++i )) ; do
echo item: $i
done
Whenever I try to run this it just gives me an error message saying it expected an operand.
I am really confused on the error here and any help would be greatly appreciated!
To get your code running with minimal change, replace:
fileCount= ls -1 | wc -l
With:
fileCount=$(ls -1 | wc -l)
$(...) is called command substitution. It is what you use when you want capture the output of a command in a variable.
It is very important that there be no spaces on either side of the equal sign.
Improvements
To speed up the result, use the -U option to turn off sorting.
To prevent any attempt to display special characters, use -q.
Thus:
fileCount=$(ls -1Uq | wc -l)
Lastly, when ls is writing to something other than a terminal, such as, in this command, a pipeline, it prints one file name per line. This makes -1 optional.
you missed to assign the output of wc -l to your variable. Try this:
fileCount=$(ls | wc -l)
(option "-1" is not needed, because ls writes one file per line if its stdout is not a terminal)
in my script i need to loop through lines in a file, once i find some specific line i need to save it to variable so later on i can use it outside the loop, i tried the following but it wont' work:
count=0
res=""
python my.py -p 12345 |
while IFS= read -r line
do
count=$((count+1))
if [ "$count" -eq 5 ]; then
res=`echo "$line" | xargs`
fi
done
echo "$res"
it output nothing, i also tried this,
res=""
... in the loop...
res=$res`echo "$line" | xargs`
still nothing. please help. thanks.
Update: Thanks for all the help. here is my final code:
res=python my.py -p 12345 | sed -n '5p' | xargs
for finding a specific line in a file, have you considered using grep?
grep "thing I'm looking for" /path/to/my.file
this will output the lines that match the thing you're looking for. Moreover this can be piped to xargs as in your question.
If you need to look at a particularly numbered line of a file, consider using the head and tail commands (which can also be piped to grep).
cat /path/to/my.file | head -n5 | tail -n1 | grep "thing I'm looking for"
These commands take the first lines specified (in this case, 5 and 1 respectively) and only prints those out. Hopefully this will help you accomplish your task.
Happy coding! Leave a comment if you have any questions.
conducting a word count of a directory.
ls | wc -l
if output is "17", I would like the output to display as "017".
I have played with | printf with little luck.
Any suggestions would be appreciated.
printf is the way to go to format numbers:
printf "There were %03d files\n" "$(ls | wc -l)"
ls | wc -l will tell you how many lines it encountered parsing the output of ls, which may not be the same as the number of (non-dot) filenames in the directory. What if a filename has a newline? One reliable way to get the number of files in a directory is
x=(*)
printf '%03d\n' "${#x[#]}"
But that will only work with a shell that supports arrays. If you want a POSIX compatible approach, use a shell function:
countargs() { printf '%03d\n' $#; }
countargs *
This works because when a glob expands the shell maintains the words in each member of the glob expansion, regardless of the characters in the filename. But when you pipe a filename the command on the other side of the pipe can't tell it's anything other than a normal string, so it can't do any special handling.
You coud use sed.
ls | wc -l | sed 's/^17$/017/'
And this applies to all the two digit numbers.
ls | wc -l | sed '/^[0-9][0-9]$/s/.*/0&/'
I would like to help out my girlfriend - she needs the specific count of certain characters in around 200 files (per file).
I already found How can I use the UNIX shell to count the number of times a letter appears in a text file?, but that only shows the complete number, not the number of occurrences per file. basically, what I want is the following:
$ ls
test1 test2
$ cat test1
ddddnnnn
ddnnddnnnn
$ cat test2
ddnnddnnnn
$ grep -o 'n' * | wc -w
16
$ <insert command here>
test1 10
test2 6
$
or something similar regarding the output. As this will be on her university machine, I cannot code anything in perl or so, just shell is allowed. My shell knowledge is a bit rusty, so I cannot come up with a better solution - maybe you could be of assistance.
grep -Ho n * | uniq -c
produces
10 test1:n
6 test2:n
If you want exactly your output:
grep -Ho n * | uniq -c | while read count file; do echo "${file%:n} $count"; done
It's not exactly elegant, but the most obvious solution is:
letter='n'
for file in *; do
count=`grep -o $letter "$file" | wc -w`
echo "$file contains $letter $count times"
done
Glen's answer is far better for the flavors of UNIX that support it. This will work on a UNIX that claims it is POSIX-compliant. This is meant for the poor folks for whom the other answer does not fly.
POSIX grep says nothing about grep -H -o See: http://pubs.opengroup.org/onlinepubs/009604499/utilities/grep.html
Get a list of the files you want call it list.txt. I chose the character ^ == shift 6 for no reason
while read fname
do
cnt=`tr -dc '^' < $fname | wc -c`
echo "$fname: $cnt"
done < list.txt