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generate rows from date.

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Top Forums Shell Programming and Scripting generate rows from date.
# 8  
Old 08-22-2007
Quote:
Originally Posted by ahmedwaseem2000
or try this code:

Code:
{

split ($2, LowerDate, "-")
split ($3, UpperDate, "-")
print 
 for(i=LowerDate[0]; i<=UpperDate[0]; i++) {

NewDate= sub( LowerDate[0], i, $2 )
print $1 NewDate $3
    }
}

i have ran this is the output ,sorry for truble. it is not incrementing and not printing propely.

$> awk -f myfiletest2.awk mydtfile.txt
1111 1999-01-01 2010-01-01
111112010-01-01
2222 2004-01-01 2040-01-01
222212040-01-01
# 9  
Old 08-22-2007
Quote:
Originally Posted by GrepMe
Hi

i have tested the script it is just printing the year YYYY only, can you help print month and date as well.

awk -f script.awk input_file.txt

here is the output from my testing.

1111 1999 2010-01-01
1111 2000 2010-01-01
1111 2001 2010-01-01
1111 2002 2010-01-01

thanks in advance.
Pimmit
works fine here on Solaris' "nawk". What's your OS? Do you have 'gawk'?
# 10  
Old 08-22-2007
My os is AIX
Quote:
Originally Posted by vgersh99
works fine here on Solaris' "nawk". What's your OS? Do you have 'gawk'?
Thanks for your time.

My os is AIX
# 11  
Old 08-22-2007
I have tested with nawk as well on my OS aix.
Quote:
Originally Posted by GrepMe
Thanks for your time.

My os is AIX
I have tested with nawk as well on my OS aix ,it is printing only year.
# 12  
Old 08-22-2007
What version of AIX?
You didn't say if you had gawk or not AND/OR if you had 'nawk'.
I don't have AIX handy at the moment - so I cannot test.
According to AWK Compatibility List, AIX 5.3 has multiple 'awk-s' installed by default. You could try any one of them:
Quote:
Originally Posted by AIX 5.3
/bin/awk /bin/gawk /bin/nawk /usr/bin/awk /usr/bin/gawk /usr/bin/nawk
# 13  
Old 08-22-2007
sorry about the snafu - somhow it worked perfectly with gawk, but not with 'nawk'.
here's the modified script:
Code:
BEGIN {
    SEPdate="[ -]"
}
{
    n=split($0, tmpA, SEPdate)
    yStart=tmpA[2]
    yEnd=tmpA[n-2]
    for(i=yStart-1; i<yEnd;i++) {
        sub(i, i+1, $2)
        print
    }
}

# 14  
Old 08-22-2007
This code is working fine:

Code:
echo "1111 1999-01-01 2010-01-01" | awk '{

LowerYear=substr($2, 1, 4) 
RestDate=substr($2, 5, 6)
UpperYear=substr($3, 1, 4)
for(i=LowerYear; i<=UpperYear; i++) {
print $1 " "i RestDate " " $3 
    }
}'


Quote:
1111 1999-01-01 2010-01-01
1111 2000-01-01 2010-01-01
1111 2001-01-01 2010-01-01
1111 2002-01-01 2010-01-01
1111 2003-01-01 2010-01-01
1111 2004-01-01 2010-01-01
1111 2005-01-01 2010-01-01
1111 2006-01-01 2010-01-01
1111 2007-01-01 2010-01-01
1111 2008-01-01 2010-01-01
1111 2009-01-01 2010-01-01
1111 2010-01-01 2010-01-01
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