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Alexos 01-30-2012 05:00 AM
[bash] transfer last line to first place
 
I want to transfer last line in the file to the first place, without processing all file. Is it possible?
For example if I have
Code:
Line 1...
Line 2...
....
Line n-1...
Line n...
I want to receive
Code:
Line n...
Line 1...
Line 2...
....
Line n-1...
Or if it is impossible, how can I get
Code:
Line n...
Line 2...
...
Line n-1...

colucix 01-30-2012 05:45 AM
Maybe using something like
Code:
cat <(tail -n1 file) <(head -n-1 file)
this requires redirecting the output to a temporary file and rename it later. How many lines are in the file?

Alexos 01-30-2012 06:30 AM
Quote:
Originally Posted by colucix (Post 4588146)
How many lines are in the file?
unknown

Alexos 01-30-2012 07:10 AM
OK, solved)
Code:
size=`wc -l file.in | awk '{print $1}'`
let size--
cat <(tail -1 file.in) <(head -$size file.in) > file.out

colucix 01-30-2012 08:45 AM
Nice! Actually the size calculation is not necessary since option -n of the head command accepts negative numbers to print all but the last N lines of each file. Specifying
Code:
head -n-1
as in my example, will exclude the last line of the file independently from the total length. My concern about the number of lines was related only to performance.

Alexos 01-31-2012 04:46 AM
Great! thanks. but I found easier method again) without creating new files
Code:
total=`tail -n 1 "file"`
sed -i -e '$d' -e "1c\\$total\n" "file"

firstfire 01-31-2012 06:31 AM
Hi.

Here is even simpler solution:
Code:
$ cat infile2.txt
Line 1...
Line 2...
....
Line n-1...
Line n...
$ echo -e '$m0\nw' | ed infile2.txt
47
47
$ cat infile2.txt
Line n...
Line 1...
Line 2...
....
Line n-1...


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