please help with kshell script..

[amonamon]

yes number 2 is correct but year is not relevant as well as month..just day..but that is in point 2.

In point 1..
if I have in my file line like this:
....
..
2222228|2007|2|28
5656569|2007|1|31
...
..
and today is 31 of march 2007 I want both lines to be in my result file.

or if I have line like this:

....
..
2123221|2007|8|31
5633561|2007|9|30
...
..

and today is for example 30 november 2007 (last day in november)

then both lines should be in result file..
I tryed with some greps..but if does not results way I want..

THANK U for patience..

cheers,

[Peter Nikitka]

Hi,

your original request can be done easy. You have to keep in mind to escape the pipe symbol when it's NOT used as a syntax element (OR) in a pattern definition:

awk -v ym=$(date '+%Y|%m|%d') -F'|' 'BEGIN {split(ym,z,"|");lookfor="\\|"z[3]"$"}
$0 ~ lookfor {if($(NF-1) != z[2]) print}' inputfile

Your extended request regarding the last day of a month requires much more work.
Let's see, if I find it interesting enough to formulate it ...

mfG Peter

[Peter Nikitka · Accepted solution]

OK,

its your turn to test this :-)

- I contruct a RE which is created+set dynamically to the date input.
- You have take the year into account, because of the leap year computation for february.
- To get no output for the current month at the last day of a month is left as an exercise.


awk -v ym=$(date '+%Y|%m|%d') 'BEGIN {split(ym,z,"|");stdlook="\\|"z[3]"$"
lof[1]=31;
if(z[1]%4) lof[2]=28
else lof[2]=29
lof[3]=31
lof[4]=30
lof[5]=31
lof[6]=30
lof[7]=31
lof[8]=31
lof[9]=30
lof[10]=31
lof[11]=30
lof[12]=31
lmonstr="\\|1\\|31$"
for(i=2;i<13;i++) lmonstr=lmonstr"|\\|"i"\\|"lof[i]"$"

print "lmonstr: ",lmonstr

if(z[3]==lof[z[2]]) lookup=lmonstr
else lookup=stdlook
}
$0 ~ lookup' inputfile

mfG Peter

The test the last month computation, use
awk -v ym='2006|12|31' ...

[john korterman]

Hi,

try this, using your input file as $1...


#!/usr/bin/sh

LAST_DAY_OF_MONTH=no
THIS_MONTH=$(date +%m)
THIS_DAY=$(date +%d)
MONTH=$(echo ${THIS_MONTH#0})
DAY=$(echo ${THIS_DAY#0})

case $THIS_MONTH in
1|3|5|7|8|10|12)
if [ "$THIS_DAY" = "31" ]
then
LAST_DAY_OF_MONTH=yes
fi
;;
4|6|9|11)
if [ "$THIS_DAY" = "30" ]
then
LAST_DAY_OF_MONTH=yes
fi
;;
2)
;;
4|6|9|11)
if [ "$THIS_DAY" = "30" ]
then
LAST_DAY_OF_MONTH=yes
fi
;;
2)
if [ "$THIS_DAY" = "28" -o "$THIS_DAY" = "29" ]
then
LAST_DAY_OF_MONTH=yes
fi
;;
esac
if [ "$LAST_DAY_OF_MONTH" = "yes" ]
then
grep -E "\|2\|28$|\|1\31$|\|3\|31$|\|4\|30$|\|5\|31$|\|6\|30$|\|7\|31$|\
|8\|31$|\|9\|30$|10\|31$|11\|30$|12\|31$" $1
else
grep "|${THIS_DAY}"$ $1
fi


regards,
John K.

[amonamon]

thanks this looks OK but I have problems with this grep -E

I got this:

grep: illegal option -- E
Usage: grep -hblcnsviw pattern file . . .

[john korterman]

Hi again,

if your system does not like "grep -E", try using a number of "grep -e"s instead, e.g.:

grep -e "\|2\|28"$ -e "|\|1\31"$ -e "|\|3\|31"$ -e "|\|4\|30"$ -e "|\|5\
|31"$ -e "|\|6\|30"$ -e "|\|7\|31"$ -e "|\|8\|31"$ "|\|9\|30"$ "|10\|31"$ -e "|1
1\|30"$ -e "|12\|31$" $1


regards,
John K.