UNIX：如何使用sed / awk / grep打印文件中的特定行？

Question

I have a data in Unix which were fetched by a command and prints:

line01
line02
line03
line04
line05
line06
line07
line08
line09
line10
line11
line12

and I wanted to sort it out such that the lines 10 to 12 are above lines 1 to 9. like this:

line10
line11
line12
line01
line02
line03
line04
line05
line06
line07
line08
line09

i tried using

<command that fetches the data> | awk 'NR>=10 || NR<=9'

and

<command that fetches the data> | sed -n -e '4,5p' -e '1,3p'

but it still display in a sorted order. i'm new to unix so i don't know how to properly use awk/sed.

PS. These data are stored in a variable which will then be processed by another command. so i needed it to be sorted that way so that line 10-12 will be processed first. :)

Answer 1

Use head and tail :

$ tail -n 2 file && head -n 3 file
name4
name5
name1
name2
name3

Your awk and sed approach do not work because you are just saying: print lines number X, Y and Z, and they will do so as soon as they find any of them. If you wanted to use these tools, you would need to read the file first, storing its content, and then print it.

$ awk -v OFS="\n"  '{a[NR]=$0} END {print a[4], a[5], a[1], a[2], a[3]}' file
name4
name5
name1
name2
name3

Or even give the order as a variable:

awk -v order="4 5 1 2 3" 
      'BEGIN {split(order,lines)}
       {a[NR]=$0} 
       END {for (i=1;i<=length(lines);i++) print a[lines[i]]}' file

If you want to give the order of the lines as an argument, you can use process substitution saying awk '...' <(command) file and working with FNR/NR to distinguish between the input and the file

Or you can use - to read from stdin as first file:

echo "4 5 1 2 3" | awk 'FNR==NR {n=split($0,lines); next}
    {a[FNR]=$0}
    END {for (i=1;i<=n;i++) print a[lines[i]]}' - file

As one-liner:

$ echo "4 5 1 2 3" | awk 'FNR==NR {n=split($0,lines); next} {a[FNR]=$0} END {for (i=1;i<=n;i++) print a[lines[i]]}' - a

Answer 2

This might work for you (GNU sed):

sed '1h;2,9H;1,9d;12G' file

Replace the hold space with line 1, then append lines 2 to 9 to the hold space and delete lines 1 thru 9. Print all other lines normally but on line 12 append the lines stored in the hold space to the pattern space.

Answer 3

Using sort actually:

$ sort -r -s -k 1.5,1.5 /tmp/lines 
line10
line11
line12
line01
line02
line03
line04
line05
line06
line07
line08
line09

The -k 1.5,1.5 means I'm using only 5th character of first word for sorting. -r means reverse order and -s means stable - leaving lines that have same 5th character in the same order.