i want to get the line number from a file.In a file where 5 dashes(-----) is appear i want to get that line number.
For example
Day to Day
------------------
Month
-----
Jan
feb
mar
Year
-----
2014
2013
2012
so here Month and year have 5 dashes.i want to get that month and year line number.
Thanks.
If you only want the line number, and not the text behind, you can use cut in complement of grep:
grep -n '^-----$' file.txt | cut -f1 -d:
Output:
4
11
You can use awk:
awk '/-{5}/{print NR, $0}' file
2 ------------------
4 -----
11 -----
To get only line numbers use:
awk '/-{5}/{print NR}' file
2
4
11
EDIT: If you want to find lines with exact 5 dashes then use:
awk '/^-{5}$/{print NR}' file
4
11
One of sed's lesser-used commands:
sed -n '/^-----$/=' file
you can use the -n option with grep to make grep output the line number. So, it would be something like this:
grep -n ^-----$ filename.txt
Related
I have created a simple script that prints the contents of a text file using cat command. Now I want to print a line along with its number, but at the same time I need to ignore blank lines. The following format is desired:
1 George Jones Berlin 2564536877
2 Mike Dixon Paris 2794321976
I tried using
cat -n catalog.txt | grep -v '^$' catalog.txt
But I get the following results:
George Jones Berlin 2564536877
Mike Dixon Paris 2794321976
I have managed to get rid of the blank lines, but line's number is not printed. What am I doing wrong?
Here are the contents of catalog.txt:
George Jones Berlin 2564536877
Mike Dixon Paris 2794321976
Your solution doesn't work because cat -n catalog.txt is already giving you non-blank lines.
You can pipe grep's output to cat -n:
grep -v '^$' yourFile | cat -n
Example:
test.txt:
Hello
how
are
you
?
$ grep -v '^$' test | cat -n
1 Hello
2 how
3 are
4 you
5 ?
At first glance, you should drop the file name in the command line to grep to make grep read from stdin:
cat -n catalog.txt | grep -v '^$'
^^^
In your code, you supplied catalog.txt to grep, which made it read from the file and ignore its standard input. So you're basically grepping from the file instead of the output of cat piped to its stdin.
To correctly ignore blank lines the prepend line numbers, switch the order of grep and cat:
grep -v '^$' catalog.txt | cat -n
Another awk
$ awk 'NF{$0=FNR " " $0}NF' 48488182
1 George Jones Berlin 2564536877
3 Mike Dixon Paris 2794321976
The second line was blank in this case.
single, simple, basic awk solution could help you here.
Solution 1st:
awk 'NF{print FNR,$0}' Input_file
Solution 2nd: Above will print line number including the line number of NULL lines, in case you want to leave empty lines line number then following may help you in same.
awk '!NF{FNR--;next} NF{print FNR,$0}' Input_file
Solution 3rd: Using only grep, though output will have a colon in between line number and the line.
grep -v '^$' Input_file | grep -n '.*'
Explanation of Solution 1st:
NF: Checking condition here if NF(Number of fields in current line, it is awk's out of the box variable which has the value of number of fields in a line) is NOT NULL, if this condition is TRUE then following the actions mentioned next to it.
{print FNR,$0}: Using print function of awk here to print FNR(Line number, which will have the line's number in it, it is awk's out of box variable) then print $0 which means current line.
By this we satisfy OP's both the conditions of leaving empty lines and print the line numbers along with lines too. I hope this helps you.
This question already has answers here:
Sort a text file by line length including spaces
(13 answers)
Closed 9 years ago.
I'm new to linux and checking to see if there is any sort command available for below scenario. I have a file containing lines like below.
this is a 10
this is 5
this 40
this is a boy in 3
this is a boy 6
I would like to sort it like
this 40
this is 5
this is a 10
this is a boy 6
this is a boy in 3
Any help is appreciated!
Assuming all of your data is in a text file called testfile this answer should work:
cat testfile | awk '{ print length, $0 }' | sort -n | cut -d" " -f2-
It comes from this question: Sort a text file by line length including spaces
If you don't care about blank lines you can run
$ sort <filename>
this 40
this is 5
this is a 10
this is a boy 6
this is a boy in 3
sort can read a file and organize the output. If you want to remove the empty lines, one option you have is sed
$ sort test.txt | sed '/^$/d'
this 40
this is 5
this is a 10
this is a boy 6
this is a boy in 3
I have output from a customised log file like this:
8 24 yum
8 24 yum
8 24 make
8 24 make
8 24 cd
8 24 cd
8 25 make
8 25 make
8 25 make
8 26 yum
8 26 yum
8 26 make
8 27 yum
8 27 install
8 28 ./linux
8 28 yum
I'd like to know if there's anyway to count the number of specific values of the third field. For example I may want to count the number of cd,yum and install only.
You can use awk to do get the third field values and wc -l to count the number.
awk '$3=="cd"||$3=="yum"||$3=="install"||$3=="cat" {print $0}' file | wc -l
You can also use egrep, but this will look for these words not only on the third field, but everywhere else in the line.
egrep "(cd|yum|install|cat)" file | wc -l
if you want to count a specific word on the third field, then you can do the above without multiple regexs.
awk '$3=="cd" {print $0}' | wc -l
A classic shell script to do the job is:
awk '{print $3}' "$file" | sort | uniq -c | sort -n
Extract values from column 3 with awk, sort the identical names together, count the repeats, sort the output in increasing order of count. The sort | uniq -c | sort -n part is a common meme.
If you're using GNU awk, you can do it all in the awk script; it might be more efficient, but for really humungous files, it can run out of memory where the pipeline doesn't (sort spills to disk when necessary; writing code to spill to disk in awk is not sensible).
Use cut, sort and uniq:
$ cut -d" " -f3 inputfile | sort | uniq -c
2 cd
1 install
1 ./linux
6 make
6 yum
For your input this
awk '{++a[$3]}END{for(i in a)print i "\t" a[i];}' file
Would print:
cd 2
install 1
./linux 1
make 6
yum 6
Using awk to count the occurrences of field three and sort to order the output:
$ awk '{a[$3]++}END{for(k in a)print a[k],k}' file | sort -n
1 install
1 ./linux
2 cd
6 make
6 yum
So filter by command:
$ awk '/cd|yum|install/{a[$3]++}END{for(k in a)print a[k],k}' file | sort -n
1 install
2 cd
6 yum
To stop partial matches such as grep in egrep use word boundaries \< and \> so the filter would be /\<cd\>|\<yum\>|\<install\>/
You can use grep to filter by multiple terms at the same time:
cut -f3 -d' ' file | grep -x -e yum -e make -e install | sort | uniq -c
Explanation:
The -x flag is to match only the lines that match exactly, as if with ^pattern$
The cut extracts the 3rd column only
We sort, uniq with count in the end for efficiency, after all junk is removed from the input
i guess u want to count the values of yum install & cd separately. if so, u shud go for 3 separate awk statements: awk '$3=="cd" {print $0}' file | wc -l
awk '$3=="yum" {print $0}' file | wc -l
awk '$3=="install" {print $0}' file | wc -l
Hello I have a problem in bash.
i have a file and i am trying insert a point in the final line of each line:
cat file | sed s/"\n"/\\n./g > salida.csv
but not works =(.
Because i need count the lines with a word
I need count the lines with the same country
and if i do a grep the grep take colombia and colombias.
And other question how i can count lines with the same country?
for example
1 colombia
2 brazil
3 ecuador
4 colombias
5 colombia
colombia 2
colombias 1
ecuador 1
brazil 1
how about
cut -f2 -d' ' salida.csv | sort | uniq -c
since a sed solution was posted (probably the best tool for this task), I'll contribute an awk
awk '$NF=$NF"."' file > salida.csv
Update:
$ cat input
1 colombia
2 brazil
3 ecuador
4 colombias
5 colombia
$ awk '{a[$2]++}END{for (i in a) print i, a[i]}' input
brazil 1
colombias 1
ecuador 1
colombia 2
...and, please stop updating your question with different questions...
Your command line has a few problems. Some that matter, some that are style choices, but here's my take:
Unnecessary cat. sed can take a filename as an argument.
Your sed command doesn't need the g. Since each line only has one end, there's no reason to tell it to look for more.
Don't look for the newline character, just match the end of line with $.
That leaves you with:
sed s/$/./ file > salida.csv
Edit:
If your real question is "How do I grep for colombia, but not match colombias?", you just need to use the -w flag to match whole words:
grep -w colombia file
If you want to count them, just add -c:
grep -c -w colombia file
Read the grep(1) man page for more information.
let's have
126 Mar 8 07:45:09 nod1 /sbin/ccccilio[12712]: INFO: sadasdasdas
2 Mar 9 08:16:22 nod1 /sbin/zzzzo[12712]: sadsdasdas
1 Mar 8 17:20:01 nod1 /usr/sbin/cron[1826]: asdasdas
4 Mar 9 06:24:01 nod1 /USR/SBIN/CRON[27199]: aaaasdsd
1 Mar 9 06:24:01 nod1 /USR/SBIN/CRON[27201]: aaadas
I would like to sort this output by date and time key.
Thank you very much.
Martin
For GNU sort: sort -k2M -k3n -k4
-k2M sorts by second column by month (this way "March" comes before "April")
-k3n sorts by third column in numeric mode (so that " 9" comes before "10")
-k4 sorts by the fourth column.
See more details in the manual.
little off-topic - but anyway. only useful when working within filetrees
ls -l -r --sort=time
from this you could create a one-liner which for example deletes the oldest backup in town.
ls -l -r --sort=time | grep backup | head -n1 | while read line; do oldbackup=\`echo $line | awk '{print$8}'\`; rm $oldbackup; done;
days need numeric (not lexical) sort, so it should be sort -s -k 2M -k 3n -k 4,4
See more details here.
You can use the sort command:
cat $logfile | sort -M -k 2
That means: Sort by month (-M) beginning from second column (-k 2).