Good day to all,
I was wondering how to find the line number of a line with only commas. The only but is that I don't know how many commas have each line:
Input:
...
Total,Total,,,
,,,,
,,,,
Alemania,,1.00,,
...
Thanks in advance for any clue
You can do this with a single command:
egrep -n '^[,]+$' file
Line numbers will be prefixed.
Result with your provided four test lines:
2:,,,,
3:,,,,
Now, if you only want the line numbers, you can cut them easily:
egrep -n '^[,]+$' file | cut -d: -f1
sed
sed -n '/^,\+$/=' file
awk
awk '/^,+$/&&$0=NR' file
With GNU sed:
sed -nr '/^,+$/=' file
Output:
2
3
Related
I have to write a function taking as argument a csv file named players.csv and a number giving the line to print.
Indded, i have to print the nth line 2column and 3rd column with a "is" between. For example Mike is John. column delimeter is ";".
I have the following code which is working :
sed -n "$2p" players.csv | cut -d ";" -f 2,3 --output-delimiter=' is '
However, I have to do the same without using cut. I can only use sed and wc. Do you have any idea what sed command I can use to have the same behavior as with cut.
Thank you for your attention and your help.
You were almost there:
sed -En "$2"'s/[^;]*;([^;]*);([^;]*).*/\1 is \2/p' players.csv
I have been trying to solve a simple sed line deletion problem.
Looked here and there. It didn't solve my problem.
My problem could simply be achieved by using sed -i'{/^1\|^2\|^3/d;}' infile.txt which deletes lines beginning with 1,2 and 3 from the infile.txt.
But what I want instead is to take the starting matching patterns from a file than manually feeding into the stream editor.
E.g: deletePattern
1
3
2
infile.txt
1 Line here
2 Line here
3 Line here
4 Line here
Desired output
4 Line here
Thank you in advance,
This grep should work:
grep -Fvf deletePattern infile.txt
4 Line here
But this will skip a line if patterns in deletePattern are found anywhere in the 2nd file.
More accurate results can be achieved by using this awk command:
awk 'FILENAME == ARGV[1] && FNR==NR{a[$1];next} !($1 in a)' deletePattern infile.txt
4 Line here
Putting together a quick command substitution combined with a character class will allow a relatively short oneliner:
$ sed -e "/^[$( while read -r ch; do a+=$ch; done <pattern.txt; echo "$a" )]/d" infile.txt
4 Line here
Of course, change the -e to -i for actual in-place substitution.
With GNU sed (for -f -):
sed 's!^[0-9][0-9]*$!/^&[^0-9]/d!' deletePattern | sed -f - infile.txt
The first sed transforms deletePattern into a sed script, then the second sed applies this script.
Try this:
sed '/^[123]/ d' infile.txt
Is it possible, in UNIX, to print a particular line of a file? For example I would like to print line 10 of file example.c. I tried with cat, ls, awk but apparently either these don't have the feature or I'm not able to properly read the man :-).
Using awk:
awk 'NR==10' file
Using sed:
sed '10!d' file
sed -n '10{p;q;}' example.c
will print the tenth line of example.c for you.
Try head and tail, you can specify the amount of lines and where to start.
To get the third line:
head -n 3 yourfile.c | tail -n 1
head -n 10 /tmp/asdf | tail -n 1
Unfortunately, all other solutions which use head/tail will NOT work incorrectly if line number provided is larger than total number of lines in our file.
This will print line number N or nothing if N is beyond total number of lines:
grep "" file | grep "^20:"
If you want to cut line number from output, pipe it through sed:
grep "" file | grep "^20:" | sed 's/^20://'
Try this:
cat -n <yourfile> | grep ^[[:space:]]*<NUMBER>[[:space:]].*$
cat -n numbers the file
the regex of grep searches the line numbered ;-)
The original mismatched as mentioned in the comments.
Te current one looks for the exact match.
- i.e. in the particular cas we need a line starting with an arbitrary amount () of spaces the followed by a space followed by whatever (.)
In case anyone thumbles over this regex and doesn't get it at all - here is a good tutorial to get you started: http://regex.learncodethehardway.org/book/ (it uses python regex as an example tough).
This might work for you:
sed '10q;d' file
I know this is basic, but I couldn't find the simplest way to iterate through a file with hundreds of lines and extract a substring.
If I have a file:
ABCY uuuu
UNUY uuuu
...
I want to end up with:
uuuu
uuuu
....
Ideally do a substring
{5} detect at character 5 and output that
You need no sed:
cut -c5-9 yourfile
It would be easier to use cut or awk. Assuming that your fields are separated by a space and you want the second field, you can use:
cut -d' ' -f2 file.txt
awk '{print $2}' file.txt
You can also use cut and awk to extract substrings:
cut -c6- file.txt
awk '{print substr($0,6);}' file.txt
However, if you really want to iterate through the file and extract substrings, you can use a while loop:
while IFS= read -r line
do
echo ${line:5}
done < file.txt
if you really love sed, you could try:
sed -r 's/^.{5}//' file
I have a sql dump with 300mb that gives me an error on specific line.
But that line is in the middle of the file. What is the best approach?
head -n middleLine dump.sql > output?
Or can i output only the line i need?
You could use sed -n -e 123456p your.dump to print line 123456
If the file is long, consider using
sed -n 'X{p;q}' file
Where X is the line number. It will stop reading the file after reaching that line.
If sed is too slow for your taste you may also use
cat $THE_FILE | head -n $DESIRED_LINE | tail -n 1
You can use sed:
sed -n "x p" dump.sql
where x is the line number.
This might work for you:
sed 'X!d;q' file
where X is the line number.
This can also be done with Perl:
perl -wnl -e '$. == 4444444 and print and exit;' FILENAME.sql
4444444 being the line number you wish to print.
You can also try awk like:
awk 'NR==YOUR_LINE_NO{print}' file_name
If you know a phrase on that line I would use grep. If the phrase is "errortext" use:
$ cat dump.sql | grep "errortext"