How to count the number of lines in a text file starting with a certain word?
I do not want to use sed and then wc -l. Any better solution?
Just grep your word and then use wc -l to count the lines... like this
grep '^your_word' /path/to/file | wc -l
Try this:-
awk '/^yourwordtofind/{a++}END{print a}' file
grep -c "pattern" <filename>
For example: If you want to search for a pattern more in a file test.txt, then below is the command:
grep -c "more" test.txt
cat <filename> | grep <word> | wc -l
Related
I've got a command to perform a series of commands that produce a variable output string such as 123456. I want to pipe that to a sed command replacing a known string in a csv file that looks like this:
Fred,Wilma,Betty,Barney
However, the command below does not work and I haven't found any other references to using pipe values as the variable for a replace.
How does this code change if the values in the csv are in a random order and I always want to change the second value?
Example code:
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt |
sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
Contents of input.csv should become: Fred,0d522cd316,Betty,Barney
Okay, in
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt | sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
you have a bug. That "> test.txt" after cut is going to eat your stdin on sed, so things go weird with that pipe afterwards taking stdin. You don't want a pipe there, or you don't want to redirect to a file.
The way to take piped stdin and use it as a parameter in a command is through xargs.
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 | xargs --replace=INSERTED -- sed -i -e 's/Wilma/INSERTED/' input.csv
(...though that find|shasum is suspect too, in that the order of files is random(ish) and it matters for a reliable sum. You prpobably mean to "|sort" after find.)
(Some would sed -i -e "s/Wilma/$(find|sort|shasum|cut)" f, but I ain't among them. Animals.)
For replacing a fixed string like "Wilma", try:
sed -i 's/Wilma/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/' input.csv
To replace the 2nd field no matter what's in it, try:
sed -i 's/[^,]*/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/2' input.csv
I like to a grep a character from a string then count it, I don't see from google search. Please advise. I might miss search for it.
node_count=`echo "test1|test2|test3" | grep "|" |wc -l`|echo $node_count
output is always 1 to me.
1
Remember that I don't grep from a file but a line of string. Grep from a file is easy.
You might want to use option -o of grep:
$ node_count=`echo "test1|test2|test3" | grep "|" -o |wc -l` && echo $node_count
# 2
If we use grep -c option it will give you the each occurrences only once per line. But I need the total no of matched occurrences not line count.
Use this
grep -o pattern path | wc -l
You can use -o flag to output only the matched part and then pipe it to wc -w to get word count.
Eg: ls ~ | grep -o pattern | wc -w
I have bunch of files in a directory,
I need to change prefix of lines in file like "AB_" to "YZ_"
how can i do it?
i have used grep and sed like,
grep -nr "AB_" ./ | xargs -0 sed -i 's/AB_/YZ_/g'
but giving error,
: File name too long
example string in a file are: Hello AB_WORLD! and Hello WORLD_AB!
Thanks.
sed will take multiple files as arguments, so this should work:
sed -i '/AB_/s//YZ_/g' *
(Note that -i is non-standard)
You mean grep -lr not grep -nr
-l gives you the file name; -n gives you the matching line with line number prepended
I like Perl for this one:
The -i option will save the original file with a.bak extension.
$ perl -i.bak -pe 's/^AB_/YZ_/' *.txt
grep -lr "AB_" ./ | while read file
do
echo "Change file $file ..."
sed -i 's/AB_/YZ_/g' ${file}
done
sed one-liner answer
Find php files in the directory containing string "foo" and replace all occurences with "bar"
grep -l foo *.php | xargs sed -i '' s/foo/bar/g
To recurse through directories
grep -rl foo * | xargs sed -i '' s/foo/bar/g
(just done successfully on 8100 files)
grep -rl bar * | wc -l
8102
I have a text file and I need to search that file and figure how many blank lines are in the file. A blank line is a line with no characters.
I must use egrep.
[aman#localhost ~]$ cat >try
sldjjsd
dkfjkjdf
dfkjdf
[aman#localhost ~]$ egrep '^$' try|wc -l
4
This will do.
egrep '^$' blankfile -c
Another way, without egrep.
echo $(($(cat blank | wc -l)-$(cat blank | tr -s "\n" | wc -l)))