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04-17-2003
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Julian Date
I have a shell script which gets passed a parameter which is a combination of Year and Julian Date <YYYYj>. So April 11th, julian date is 101. So if I wanted April 11th for 2003 I would get the following value 2003101. How would I convert that in unix to be 20030411? I am using the korn shell.
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04-17-2003
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Well if you can alter the script that's passing the parameter, you could change the piece of code, such as date +%Y%j, so that it sends what you want it to send, such as date +%Y%m%d.
Otherwise... Assuming the variable that stores that passed value is called varPassed, you can put this (ksh) code in a file called someScript, make it executable, and call it with newValue=`someScript $varPassed`: Code:
varPassed=$1
month=1
flag=0
theYear=`echo $varPassed | awk '{print substr ($0, 1, 4)'}`
dayOfYear=`echo $varPassed | awk '{print substr ($0, 5, 3)'}`
dayOfYear=`expr $dayOfYear - 0`
TestLeapYear() {
if [[ `expr $theYear % 4` != 0 && $dayOfYear -gt 28 ]] then
month=`expr $month + 1`
dayOfYear=`expr $dayOfYear - 28`
else
if [[ `expr $theYear % 4` = 0 && $dayOfYear -gt 29 ]] then
month=`expr $month + 1`
dayOfYear=`expr $dayOfYear - 29`
else
flag=1
fi
fi
}
Days30() {
if [[ $dayOfYear -gt 30 ]] then
month=`expr $month + 1`
dayOfYear=`expr $dayOfYear - 30`
else
flag=1
fi
}
Days31() {
if [[ $dayOfYear -gt 31 ]] then
month=`expr $month + 1`
dayOfYear=`expr $dayOfYear - 31`
else
flag=1
fi
}
Days31 #january
if [[ $flag = 0 ]] then
TestLeapYear #february
fi
if [[ $flag = 0 ]] then
Days31 #march
fi
if [[ $flag = 0 ]] then
Days30 #april
fi
if [[ $flag = 0 ]] then
Days31 #may
fi
if [[ $flag = 0 ]] then
Days30 #june
fi
if [[ $flag = 0 ]] then
Days31 #july
fi
if [[ $flag = 0 ]] then
Days31 #august
fi
if [[ $flag = 0 ]] then
Days30 #september
fi
if [[ $flag = 0 ]] then
Days31 #october
fi
if [[ $flag = 0 ]] then
Days30 #november
fi
if [[ $month -lt 10 ]] then
month=0$month
fi
if [[ $dayOfYear -lt 10 ]] then
dayOfYear=0$dayOfYear
fi
echo $theYear$month$dayOfYear
Last edited by oombera; 04-17-2003 at 11:52 AM.
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04-17-2003
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What you're doing is more of a day-of-year thing rather than a Julian number thing. But a Julian number calculator can handle this stuff easily.
Get the Julian number for the day before Jan 1 of the year in question: datecalc -j 2002 12 31 52639 Now calculate the JD of the date you want: 52740=52639+101 Lastly, convert that to a date: datecalc -j 52740 2003 4 11 Of course, you will want to read the output of the last command into some variables, not just display them. With ksh, this is really just a one-liner: Code:
Y=2003 DOY=101 datecalc -j $(($(datecalc -j $((Y-1)) 12 31) + DOY)) | read year month day I see oombera beat me to the punch here 8) Oh well, now you have two options..
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