Just wondering if you could lend me a little hand. Basically I want to
rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
(ignore the date) should be the last written time, so far I've got to
this stage:
stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
so I get :
11:01:09.000000000
How can I get rid of the leading 0'swithout having to pipe the output
to anotehr awk statement, is it possible to do this withing the
current awk statement?
Thanks
Dan
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08-20-2008, 10:11 AM
Rui Miguel Silva Seabra
Awk question
On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
> Just wondering if you could lend me a little hand. Basically I want to
> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
> (ignore the date) should be the last written time, so far I've got to
> this stage:
>
> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
>
> so I get :
> 11:01:09.000000000
>
> How can I get rid of the leading 0'swithout having to pipe the output
> to anotehr awk statement, is it possible to do this withing the
> current awk statement?
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08-20-2008, 10:58 AM
Chris G
Awk question
On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
> Guys,
>
> Just wondering if you could lend me a little hand. Basically I want to
> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
> (ignore the date) should be the last written time, so far I've got to
> this stage:
>
> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
>
> so I get :
> 11:01:09.000000000
>
> How can I get rid of the leading 0'swithout having to pipe the output
> to anotehr awk statement, is it possible to do this withing the
> current awk statement?
>
Probably, yes, look at the substr() function.
You don't need the sed in the pipe either I don't think, you could do
this with the pattern matching capabilities of awk, e.g. (not tested) :-
stat log | awk -F ' ' '/Modify:/ {print $3}'
(N.B. I haven't done the substr() bit)
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08-20-2008, 01:20 PM
"Patrick O'Callaghan"
Awk question
On Wed, 2008-08-20 at 11:03 +0100, Dan Track wrote:
> Guys,
>
> Just wondering if you could lend me a little hand. Basically I want to
> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
> (ignore the date) should be the last written time, so far I've got to
> this stage:
>
> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
>
> so I get :
> 11:01:09.000000000
>
> How can I get rid of the leading 0'swithout having to pipe the output
> to anotehr awk statement, is it possible to do this withing the
> current awk statement?
Use printf instead of print.
poc
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08-21-2008, 08:14 AM
"Dan Track"
Awk question
On Wed, Aug 20, 2008 at 11:11 AM, Rui Miguel Silva Seabra <rms@1407.org> wrote:
> On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
>> Just wondering if you could lend me a little hand. Basically I want to
>> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
>> (ignore the date) should be the last written time, so far I've got to
>> this stage:
>>
>> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
>>
>> so I get :
>> 11:01:09.000000000
>>
>> How can I get rid of the leading 0'swithout having to pipe the output
>> to anotehr awk statement, is it possible to do this withing the
>> current awk statement?
>
> There simpler solutions but this works:
>
> stat log | awk '/Modify/ { print $3 }' | cut -d . -f 1
>
Hi
I had already changed it to do it the way you mentioned. Guess there's
no way to do a second break within AWK.
Thanks Guys,
Dan
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08-21-2008, 02:08 PM
"Patrick O'Callaghan"
Awk question
On Thu, 2008-08-21 at 09:14 +0100, Dan Track wrote:
> On Wed, Aug 20, 2008 at 11:11 AM, Rui Miguel Silva Seabra <rms@1407.org> wrote:
> > On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
> >> Just wondering if you could lend me a little hand. Basically I want to
> >> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
> >> (ignore the date) should be the last written time, so far I've got to
> >> this stage:
> >>
> >> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
> >>
> >> so I get :
> >> 11:01:09.000000000
> >>
> >> How can I get rid of the leading 0'swithout having to pipe the output
> >> to anotehr awk statement, is it possible to do this withing the
> >> current awk statement?
> >
> > There simpler solutions but this works:
> >
> > stat log | awk '/Modify/ { print $3 }' | cut -d . -f 1
> >
>
>
> Hi
>
> I had already changed it to do it the way you mentioned. Guess there's
> no way to do a second break within AWK.
As I said before, use printf:
stat log | awk '/Modify/ { printf "%.8s
", $3 }'
poc
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08-21-2008, 03:17 PM
Roger Heflin
Awk question
Dan Track wrote:
On Wed, Aug 20, 2008 at 11:11 AM, Rui Miguel Silva Seabra <rms@1407.org> wrote:
On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
Just wondering if you could lend me a little hand. Basically I want to
rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
(ignore the date) should be the last written time, so far I've got to
this stage:
stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
so I get :
11:01:09.000000000
How can I get rid of the leading 0'swithout having to pipe the output
to anotehr awk statement, is it possible to do this withing the
current awk statement?
I had already changed it to do it the way you mentioned. Guess there's
no way to do a second break within AWK.
Thanks Guys,
Dan
It's trivial, you need to use the split function.
stat log | awk '/Modify/ { split($3,parts,".") ;print parts[1] }'
Roger
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08-21-2008, 10:03 PM
"NiftyFedora Mitch"
Awk question
On Thu, Aug 21, 2008 at 8:17 AM, Roger Heflin <rogerheflin@gmail.com> wrote:
> Dan Track wrote:
>> On Wed, Aug 20, 2008 at 11:11 AM, Rui Miguel Silva Seabra <rms@1407.org>
>> wrote:
>>> On Wed, Aug 20, 2008 at 11:03:58AM +0100, Dan Track wrote:
>>>>
>>>> Just wondering if you could lend me a little hand. Basically I want to
>>>> rename a file from log.1 log.2 etc to log.10.36.34. The time stamp
>>>> (ignore the date) should be the last written time, so far I've got to
>>>> this stage:
>>>>
>>>> stat log | sed -n '/Modify:/p' | awk -F ' ' '{print $3}'
>>>>
>>>> so I get :
>>>> 11:01:09.000000000
>>>>
>>>> How can I get rid of the leading 0'swithout having to pipe the output
>>>> to anotehr awk statement, is it possible to do this withing the
>>>> current awk statement?
>>>
>>> There simpler solutions but this works:
>>>
>>> stat log | awk '/Modify/ { print $3 }' | cut -d . -f 1
>>>
>>
>>
>> Hi
>>
>> I had already changed it to do it the way you mentioned. Guess there's
>> no way to do a second break within AWK.
>>
>> Thanks Guys,
>>
>> Dan
>>
> It's trivial, you need to use the split function.
>
> stat log | awk '/Modify/ { split($3,parts,".") ;print parts[1] }'
>
>
Awk question
