如何使用Bash（sed或任何其他）命令在EOF-nth行插入

Question

I did search in Stack Overflow, but, I couldn't find a solution. Did try enough.

I have a file containing code like this-

...
{
...

 void abc()
 {
  if(condition)
  {
   x=y;
   z=0;
  }
 }
} //EOF

I want to insert some lines of code in between x=y; and z=0; .

This can be done by 2 methods-

1. Searching for the pattern x=y; and inserting later

2. Finding no. of lines in file -

   NUMOFLINES = $(wc -l some.txt | awk '{print $1}') TOINSERTLINE=NUMOFLINES-5

So, inserting at TOINSERTLINE's value.

More optimal methods are welcome. Help me fix this.

---

Concluded Answer from the discussion-

sed -i.bak -e '^/x=y;/r file-to-insert' *.xaml

This worked for me!!

Thanks to Jonathan Leffler and anubhava . These were the closest to what I wanted.

Answer 1

If you want to insert a line after 5th line use this sed :

sed -i.bak $TOINSERTLINE'r data-file' file

Here data to insert comes from data-file

Answer 2

This is one way:

(
    sed '/x=y;/q'
    echo "NUMOFLINES - 5"
    cat
) < file

Answer 3

awk '1; /x=y/{print "foo"}' file

awk -v nl=$(wc -l < file) '1; NR==(nl-5){ print "foo" }' file

If those don't do what you want, edit your question to clarify your requirements.