How to find missing dates?

[cuervo11]

Hello guys, I have a problem, I need find the missing dates of the year, for example this is my input:
2014-11-25T12:00:00
2014-11-26T12:00:00
2014-11-27T12:00:00
2014-11-28T12:00:00
2014-11-29T12:00:00
2014-11-30T12:00:00
2014-11-01T12:00:00
2014-11-02T12:00:00
2014-11-03T12:00:00
2014-11-04T12:00:00
2014-11-05T12:00:00
2014-11-06T12:00:00
2014-11-07T12:00:00
2014-11-08T12:00:00
2014-11-09T12:00:00
2014-11-10T12:00:00
2014-11-11T12:00:00

And the output will be all dates in that format less the dates I have, any ideas?

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[feherke]

Hi

The easiest is to load all those data in an array, then loop from January 1[sup]st[/sup] to December 31[sup]st[/sup] and display those not in the array :

gawk 'd[teal][[/teal][navy]$1[/navy][teal]]=[/teal][purple]0[/purple][teal];[/teal][b]END[/b][teal]{[/teal][navy]t[/navy][teal]=[/teal][COLOR=FF6600]mktime[/color][teal]([/teal]year[i][green]" 01 01 12 00 00"[/green][/i][teal]);[/teal][b]while[/b][teal](([/teal][navy]s[/navy][teal]=[/teal][COLOR=FF6600]strftime[/color][teal]([/teal][i][green]"%FT%T"[/green][/i][teal],[/teal]t[teal]))&&[/teal][b]substr[/b][teal]([/teal]s[teal],[/teal][purple]1[/purple][teal],[/teal][purple]4[/purple][teal])==[/teal]year[teal]){[/teal][b]if[/b][teal](!([/teal]s [b]in[/b] d[teal]))[/teal][b]print[/b] s[teal];[/teal]t[teal]+=[/teal][purple]60[/purple][teal]*[/teal][purple]60[/purple][teal]*[/teal][purple]24[/purple][teal]}}[/teal]' year=2014 /path/to/input

The above code works with GNU [tt]awk[/tt] only. You have to specify the year as variable assignment in the command line and specify the input file name too.

Note that you may want to hard code the hour of day as 12 in case your machine's time zone supports daylight saving and you not want the noon at summer to be 13:00:00. ( I mean, just change "%T" to "12:00:00". )


Feherke.
feherke.ga