how to replace certain patterns in a file thru unix!!

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# 1  
Old 08-29-2001
Question how to replace certain patterns in a file thru unix!!

Hi Friends,
well what i want to do is that suppose i have a file say like
okok i agree this is a ridiculous data for a file.
now i want that the characters in each line fromposition 16 to say 19 gets replaced by a particular string , in this case the charcters are cool pool and tool , so say i want is the follwoing output:


or maybe it is possible that the case is that i want to replcae cool,tool and pool by word whose length is greate than the ones which are present say like


and also suppose i want say that a particular word occurence in every line is replaced by some other word say like i have input file containing


and i want the file to be manipulated as

or say different lengths replacement like

the bottomline being that there r 2 cases one being replacement being on base of position and the other being replaxcement on basis of certain pattern.

HOW do i do this????
PLzzz help

# 2  
Old 08-29-2001
pattern replacement

Sounds like you need "sed".
Check out...
# 3  
Old 08-29-2001
aye sed does soudn liek the way you want to go.

sed 's/.ool/&ltreplacement string&gt/'
# 4  
Old 08-30-2001
Thanks a lot pals , one of the problems is solved but still one exists.
i am unable to replace based on positions,
could u shed some light on that part plzzzz.
# 5  
Old 08-30-2001
you will want to look into sed. we gave you the key to open the door. you just have to turn the knob.

you will want to look into regular exprestions also.
# 6  
Old 08-31-2001

Hi optimus,
i turned the knob!
looked in regular expresions too, but all i cud find is pattern replacement but noting on positional replacement!
Help if u can plzzz.
# 7  
Old 08-31-2001
A few patterns to experiment with:

Some hints:
. is any character
......  is any 6 characters
.\{6\}  also any 6 characters
\(.\{6\}\) any 6 characters saved as subexpression
^\(.\{6\}\) first 6 characters saved as subexpression

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