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awk to replace second occurance

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# 1  
Old 03-04-2010
awk to replace second occurance
Code:
#original file
.
.
~
~
Index=2  
xxx
replace  #dont replace 1st occurance
yyy
Index=2
xxx
replace  #substitue replace with "REPLACE"
yyy
Index=2
xxx
replace #substitue replace with "REPLACE"
yyy
Index=3
xxx
replace 
yyy

Code:
Expected output file:

.
.
~
~
Index=2
xxx
replace  
yyy
Index=2
xxx
REPLACE  
yyy
Index=2
xxx
REPLACE  
yyy
Index=3
xxx
replace 
yyy


Hi,

This is the simiplified file I am working on.. I need to :
1. search for 2nd occurance of a string pattern ("Index=2"in file)
2. Go forward 2 lines after this
3. Replace current line with new pattern (ex: "REPLACE" )



I am using awk for checking the 2nd oocurance, but dont know
how to step 2 lines forward and replace the string.

Code:
#awk.sh

/Index=2/{
   mcount++
   if(mcount == 2){
      {print NR+2}
      #need to substitute NR+2
   }
}

# 2  
Old 03-04-2010
Code:
awk '/^replace/ {c++}; (c==2)||(c==3) {$1=toupper($1)}1' urfile

# 3  
Old 03-04-2010
Try this:

Code:
awk '/Index=2/{print;getline;print;getline;if (cnt == 0)cnt+=1; else { gsub("replace","REPLACE")}}1' filename


cheers,
Devaraj Takhellambam
# 4  
Old 03-04-2010
Thanks guys...both the solutions solves my problem..
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