Script to replace lines in ksh Script

Shell Programming and Scripting
1

Hi All,

I am novice to Unix and I need your expert advice for the below task.

There is a KSH script file in which I need to replace few line as per the below expectations. So my file look like as

# Host Setup Command:
Line 1
Line 2
Line 3
Line 4
Line Any
_AB_Proxy_DIR=Out_ABC_For_Bussiness.xml-$$

So in between of # Host Setup Command: and _AB_Proxy_DIR=Out_ABC_For_Bussiness.xml-$$ there can be N number of lines and I need to replace those lines as per below format

# Host Setup Command:
Syntax 1
Syntax 2
Syntax 3
_AB_Proxy_DIR=Out_ABC_For_Bussiness.xml-$$

Please let me know if you experts can help me here.

Thanks in Advance.

Rupali

2

Is this homework/classwork?

3

This is the solution I need to implement in Production at one shot.

not sure about the class/home work.

4

How is Syntax 1 etc. supplied? In a file? Calculated? A constant pattern?

5

Syntax 1 is a simple text I need to replace with the particular line.

6

Try:

sed s,"Syntax 1","Particular Line",g "$WORKFILE"

When it looks as you want it to, add -i before the workfile.

hth

7

The following works with without external programs like sed.
It deletes the block between start marker and end marker, and where it's at the end marker puts a replacement text.
Each iteration of the while loop reads a line from "./file" into the variable $line.
The variable $block is non-null if inside the block.
The output in the while loop is redirected to a new file, that is finally copied back to "./file".

block=""
f="./file"
while IFS= read -r line
do
  if [ -n "$block" ] && [ "$line" = '_AB_Proxy_DIR=Out_ABC_For_Bussiness.xml-$$' ]
  then
    block=""
    printf "%s" "\
Syntax 1
Syntax 2
Syntax 3
"
  fi
  if [ -z "$block" ]
  then
    printf "%s\n" "$line"
    [ "$line" = '# Host Setup Command:' ] && block=1
  fi
done < "$f" > "$f.new"
[ -s "$f.new" ] &&
cp "$f.new" "$f" &&
rm "$f.new"