Hi...
I need a help in using the awk command or any other solution to avoid the usage of loops.
My question is :
I have a input like this :
where the 4th field i.e 14450 is a date field which will be passed to a converter which will automatically convert to the normal date format.
(eg : date_conv(14450) will give output 12/10/2009. )
Now i am looking for a solution which will get the output in the following manner.
o/p format :
Note '-' represents hyphen.
So in my output there will be only 6 fields like :
where field 6 = v(hexadecimal value of field6 ) - hexadecimal value of field7) - (field8*10) - field9
In My input field6=320 then the hexadecimal value will be 140
if field7=450 then the hexadecimal value will be 1C2.
So my final output will be
Please help..
Thanks.
Last edited by zaxxon; 10-12-2009 at 11:45 AM..
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Hi...
I need a help in using the awk command or any other solution to avoid the usage of loops.
My question is :
I have a input like this :
where the 4th field i.e 14450 is a date field which will be passed to a converter which will automatically convert to the normal date format.
(eg : date_conv(14450) will give output 12/10/2009. )
I wouldn't trust any date conversion code that gave that as an output. Is it October 12 or December 10?
And what relation does 14450 have to either of those?
Quote:
Now i am looking for a solution which will get the output in the following manner.
o/p format :
Note '-' represents hyphen.
So in my output there will be only 6 fields like :
where field 6 = v(hexadecimal value of field6 ) - hexadecimal value of field7) - (field8*10) - field9
In My input field6=320 then the hexadecimal value will be 140
if field7=450 then the hexadecimal value will be 1C2.
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