I want to display content of a file. The way I wanted is, when I specify a line number, it should show only that row. For example the content in a file named "FILE" are as follows:
/home/john
/home/mathew
/home/alexander
/home/testuser
/home/hellouser
I want to display single row at a time by giving line number, like If I specify 3, it should show only following row.
/home/alexander
I know another way is possible with head and tail using -n flag, but that just display the entire rows upto the line number we specify, like as follows.
head -n3 FILE
/home/john
/home/mathew
/home/alexander
I don't want that, I want only to display "/home/alexander" only the when I choose line number 3. How is it possible?
with head and tail you can do the following:
head -3 FILE | tail -1
Some other ways (4th line):
sed -n '4p' file
or
perl -ne 'print if 4 == $.' file
or
grep -n ^ file | grep '^4:' | cut -d: -f2-
You could use awk
line=3
awk -v var="$line" 'NR==var' file
It will display the line number stored in variable line
Related
I am trying to number the lines of a txt file and hide the empty ones . I use this code :
cat -n file.txt | grep . file.txt
But it doesnt work . It ignores the cat command . I want to display all the non-empty lines and number them ( the txt file is not a static one , like a list that a user can type in ).
edit : Given the great solutions below , i would also add that grep . file.txt | cat -n also worked .
I assume you want to number the lines that remain after the empty lines are removed.
Solution #1
Use sed '/^$/d' to delete the empty lines then pipe its output to cat -n to number them:
sed '/^$/d' file.txt | cat -n
The sed program contains only one command: d (delete the line). The sed commands can be prefixed by zero, one or two addresses that tell what lines the command applies to.
In this case there is only one address /^$/. It is a regex (enclosed in /) that selects the empty lines; the lines where start of the line (^) is followed by the end of the line ($).
Solution #2
You can also use grep -v '^$' to filter out the empty lines:
grep -v '^$' file.txt | cat -n
Again, ^$ is a regular expression that matches the empty lines. -v reverses the condition and tells grep to display the lines that do not match the regex.
The commands above do not modify the file. They read the content of file.txt, process it and display the result on screen.
Update
As #robc suggests in a comment, nl is even better than cat -n to number the lines. Thank you #robc, I didn't know about nl until now (I didn't know about cat -n either). It is never too late to learn new things.
This could be easily done with awk. This will print line with line numbers and ignore empty lines.
awk 'NF{print FNR,$0}' file.txt
Explanation: Adding detailed explanation for above code.
awk ' ##Starting awk program from here.
NF{ ##Checking condition if NF(number of fields) is NOT NULL in current line then do following.
print FNR,$0 ##Printing current line number by mentioning FNR and then current line value.
}
' file.txt ##Mentioning Input_file name which we are passing to awk program here.
I would like to follow a file but tail -f always starts with the last 10 lines. Is there a way to output the entire file then follow?
My goal is to find all occurrences of a string in a log, such as tail -f streaming_log | grep "string". But also include all previous line.
I know I can do tail -f -n 10000 file but I don't want to count the lines first.
-n +<line> allows you to specify a starting line (1-based):
tail -f -n +1 file # output entire file, then wait for new content
How can I apply the following command to only a part of a text file? For example from the beginning to the line 5000.
grep "^ A : 11 B : 10" filename | wc -l
I cannot use head and then apply the above command since the text file is huge.
You could try using the sed command, which I believe does better for large files, from this question and pipe to grep.
sed -n 1,5000p file | grep ...
You can try combination of -n (prefixing each line of output with line number) and -m (limiting number of matching lines). Something like this:
grep -n -m 5000 pattern file.txt | grep -B 5000 "^5000:" | wc -l
First grep search for pattern, add line numbers and limit output to first 5000 matching lines (worst case scenario: all lines from range match pattern). Second grep match line number 5000, and print all lines before this line.
I don't know if it is more efficient solution
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
I would like to take a specific part of the last line of a file. say the values between "cursor position" 9 and 22? to take the last line I know I should use tail -1 but and the second part?
I think GNU cut can get you there. For example:
cut -c 9-22
Or, in total:
tail -1 file | cut -c 9-22
You can use the cut command.
$ echo "hello how are you doing" | cut -c 9-22
w are you doin