Shell Programming and Scripting

Hello Help,

Code:

2356798 7689867 999 000
123678 20385907 9797 666
17978975 87468976 968978 98798

I am trying to have out put which actually look for the third column value of 9797 and then it insert line there after with first, second column value exactly as the previous line and replace the third by value 5000 and the next one 700
so finally the o/p file will look like

Code:

2356798 7689867 999 000
123678 20385907 9797 666
123678 20385907 5000 700
17978975 87468976 968978 98798

and so on....

Any help on this is much appreciated.
Thanks

Please refrain from bumping up your thread and instead try to figure out how to do it yourself by using the solutions that forum'ers have provided you with...

Hi Shamrock ...I am not trying to find a shortcut, believe me.I am finding it hard to be coded.If you can , please help!Just copying the first two column of the previous line and then inserting the value of your choice to third and fourth.Its okay to put value on 3rd and 4th column but copying the first two columns from previous line is hard for me.
Any help much appreciated

You could try something like:

Code:

awk '
1
$3 == 9797 {
	$3 = 5000
	$4 = 700
	print
}' file