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Add time hours

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# 8  
Old 12-04-2015
Try "%s%02d%s\n" as the format string.
# 9  
Old 12-04-2015
Thanks RudiC

---------- Post updated at 04:56 PM ---------- Previous update was at 12:15 PM ----------

RudiC

Please help

Code:
130030005        
130030012        
130090028        
130090032

when there is a gap of more that 3hrs the code doesn't work?
# 10  
Old 12-05-2015
Can't help unless more details are presented.
# 11  
Old 12-05-2015
|RudiC

here a example
input
Code:
335193508
335203508
335213508
335223508
335233508
130030005        
130030012        
130090028        
130090032

output error
Code:
335223508
335233508
336003508
336013508
336023508
130060005
130060012

Code:
line 4: 09: value too great for base (error token is "09")

code used

Code:
while { read -N3 day; read -N2 hour; read REST; }; do printf "%s%02d%s\n" $((day+(hour>20)))  $(((hour+=3) %24)) $REST; done < file

# 12  
Old 12-05-2015
Try:
Code:
while { read -N3 day; day=${day#0}; day=${day#0}; read -N2 hour; hour=${hour#0}; read REST; }; do printf "%03d%02d%s\n" $((day+(hour>20)))  $(((hour+=3)%24)) $REST; done < file

Stripping a leading zero from hourwill fix your problem with hours 8 and 9 every day. Stripping leading zeros from day won't make any difference in December, but it will protect against problems you will run into in January through early April.

Note, however, that no attempt is made here to roll the Julian day to the next year.
Last edited by Don Cragun; 12-05-2015 at 06:12 PM.. Reason: Add note.
This User Gave Thanks to Don Cragun For This Post:
jiam912
# 13  
Old 12-06-2015
Thanks Don

---------- Post updated 12-06-15 at 01:22 AM ---------- Previous update was 12-05-15 at 05:36 PM ----------

Don
A question please, I am going to use this code always.
Then, I will have problems next year?.. with JD?
# 14  
Old 12-06-2015
Quote:
Originally Posted by jiam912
Thanks Don

---------- Post updated 12-06-15 at 01:22 AM ---------- Previous update was 12-05-15 at 05:36 PM ----------

Don
A question please, I am going to use this code always.
Then, I will have problems next year?.. with JD?
If you have timestamps greater than or equal to 21mmss on December 31st, the Julian date will be set to 366 (on non-leap years) or 367 (on leap years) instead of 001. Since your input data does not include the year, there is no way to determine whether or not the Julian day 365 is December 30th or December 31st. If your input Julian day is 366, you could add code to roll that over to 001; but if the input Julian day is 365, you have no data to determine what the result should be.
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