Replace first 7 lines of one file with content of another file

Question

I've two files. One is called file1.txt and the other is person1.txt. Both of them contain some lines of text as in the following example:

• file1.txt

  word1
  word2
  word3
  word4
  word5
  word6
  word7
  word8
  word9
  

• person1.txt

  givi sixarulidze
  

What I want to do is to replace first seven lines of code in file1.txt with the content of person1.txt.

• Desired output:

  givi sixarulidze
  word8
  word9
  

How can I achieve that?

Answer 1

cp person1.txt result.txt
tail -n+8 file1 >> result.txt

Answer 2

Using GNU sponge to do in-place editing of file1.txt:

{ cat person1.txt; tail -n +8 file1.txt; } | sponge file1.txt

This concatenates the person1.txt file with all but the 7 first lines from the file1.txt file and saves the result into file1.txt.

With sed, you would do the same with

sed -e '1r person1.txt' -e '1,7d' file1.txt | sponge file1.txt

(you could use 8,$!d in place of 1,7d), or, if using GNU sed,

sed -i -e '1r person1.txt' -e '1,7d' file1.txt

This inserts the contents of person1.txt at the start and then skips the first seven lines from the file1.txt input file.

Answer 3

Another possibility is awk:

awk 'NR==FNR || FNR>7' person1.txt file1.txt > result.txt

This will process first person1.txt and then file1.txt. It will print the current line if either

• we are processing the first of the two files, where NR (the global line-counter) is equal to FNR, the per-file line counter, or
• the per-file line-counter is larger than 7 (which will only be a limitation when processing the second file, where NR is now larger than FNR)

You even can expand this to per-file specifications of lines to be skipped:

awk -- '--skip<=0' person1.txt skip=7 file1.txt skip=4 file2.txt >result.txt

You would then simply precede each file name with a variable assignment that sets the awk variable skip to the number of lines you want to have skipped from the respective input file. The awk program checks for each processed line whether this variable, if decremented by 1, falls below zero, which would indicate that the number of lines to be skipped has been reached. The condition would then be true, and the current line printed.

Notice that the -- "end-of-options" specifier is needed because the first command of the actual program starts with a --.

Answer 4

This can be done with sed

The steps are to delete the first 6 lines. When you get to the 7th you need to read in the other file and then delete the 7th line.

sed '1,6d
7{
r person1.txt
d
} ' file1.txt > newfile1.txt &&
mv newfile1.txt file1.txt

Some versions of sed have a -i flag to do an "in place" edit.

Another way is to use tail.

{
    cat person1.txt
    tail -n +8 file1.txt
} > newfile1.txt && mv -f newfile1.txt file1.txt

Here the two commands are put together and the output redirected together.

Answer 5

$ ed file1
1,7d
0r person1
w
q

Or, as a script:

ed -s file1 <<'EOF'
1,7d
0r person1
w
q
EOF

or

printf '%s\n' 1,7d '0r person1' w q | ed -s file1

Nowadays, ex (being installed with vi) is more common than ed, although both are specified in POSIX and the latter is simpler. For real-world portability, just replace ed with ex in the code samples.