Need to find a multiple line block and replace with a multiple line block

Shell Programming and Scripting
1

I would perfer to use cut and paste to do this but I can't find a GUI to do this with. What I want to do is to find a multiple line block of code like

Exit Sub

Log_Handler:

then replace it with

GoTo RSLogRtn
Exit Sub

Log_Handler:

Basically it is just an insert, but I may want to replace whole line blocks of code at some time. How can this be accomplished. I tried using V-Grep but to no avail. It will only replace the line block with a string and you can't use expressions in the replace test.

2

can you elaborate more by giving your exact inputfile and desired output.

3

It's all there in the post...

Input: Text
OutPut: Text
Desire: Change from first text to second text...

4

Try this then...

>cat file
Exit Sub
 
Log_Handler:
 
>awk '/Exit/{err=$0;getline;getline;if($0 ~ /Log_Handler:/){print "GoTo RSLogRtn" RS err RS RS $0}}' file
GoTo RSLogRtn
Exit Sub
 
Log_Handler:
5

I will try it. Does this find/replace every reference in the file?

6

yes, if it is the same as your example. And if that's how you wanted it, redirect to another file.

7

Yes, to redirect the whole contents with the changes to another file.

---------- Post updated at 11:42 PM ---------- Previous update was at 11:37 PM ----------

Ok, now I have to find a version of AWK or GAWK that works in Windows.

Thanks.

8

what is RS err RS RS $0) here Please explian.

Thanks in advace

9

RS is for input record separator which is new line character by default.
$0 is to for all fields.
err is just a variable that assigned the value of $0.

For more information ....read

man awk