Re: please help with kshell script.. (327 Views)
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Regular Advisor
amonamon
Posts: 166
Registered: ‎05-21-2006
Message 1 of 16 (327 Views)
Accepted Solution

please help with kshell script..

hello I have file like this:


1234561|2007|2|28
4567891|2007|3|31
7897891|2007|4|5
9879871|2007|5|30
2578961|2007|6|25
...
..
.

where $2 is year $3 is month and $4 is day..

considering if today is 30may2007 I heed to have on my other file just one line:

9879871|2007|5|30

is that possible to bo done in ksh pleasee..


thanks a lot...
Honored Contributor
Peter Godron
Posts: 4,470
Registered: ‎02-13-2002
Message 2 of 16 (327 Views)

Re: please help with kshell script..

Hi,
how about:

grep -e '|2007|5|30' datafile

Honored Contributor
Peter Godron
Posts: 4,470
Registered: ‎02-13-2002
Message 3 of 16 (327 Views)

Re: please help with kshell script..

Hi (again),
or for a more generic solution:

#!/usr/bin/ksh
grep -e "$1|$2|$3" datafile

$ ./b.sh 2007 5 30
9879871|2007|5|30
Honored Contributor
john korterman
Posts: 1,117
Registered: ‎11-15-2000
Message 4 of 16 (327 Views)

Re: please help with kshell script..

Hi,

$ cat ./extract.sh

#!/usr/bin/sh
if [ $# != 4 ]
then
THIS_YEAR=$(date +%Y)
THIS_MONTH=$(date +%m)
THIS_DAY=$(date +%d)
MONTH=$(echo ${THIS_MONTH#0})
DAY=$(echo ${THIS_DAY#0})
grep "$THIS_YEAR|$MONTH|$DAY" $1
else
grep "$1|$2|$3" $4
fi


Try running it like this:

$ ./extract.sh 2007 2 28 ./infile
1234561|2007|2|28

or wait till the end of February and run it without parameters...


regards,
John K.
it would be nice if you always got a second chance
Trusted Contributor
SANTOSH S. MHASKAR
Posts: 264
Registered: ‎01-20-2004
Message 5 of 16 (327 Views)

Re: please help with kshell script..

Hi,

I assume u want to get only the line corresponding to today's date
in other file.

If so then try this

$ awk -F \| -v a=`date "+%Y"` -v b=`date "+%m"` -v c=`date "+%d"` '$2==a&&$3==b&&$4==c {print $1}' >

Thats all

-Santosh
Trusted Contributor
SANTOSH S. MHASKAR
Posts: 264
Registered: ‎01-20-2004
Message 6 of 16 (327 Views)

Re: please help with kshell script..

Hi,

Sorry for a small mistake,

Pl. replace {print $1} with {print $0}

-Santosh
Regular Advisor
amonamon
Posts: 166
Registered: ‎05-21-2006
Message 7 of 16 (327 Views)

Re: please help with kshell script..

Yes I need to have today's date and then if in file I do have some lines like this:

1111119|2007|3|31


if dodays date is april 30 I need to take that line as well as a result..

for example: today is 28 feb 2007 I need to test next date..if next date is 1.something.2007 I need to have
that line as result as well..

for example if I have:
1111119|2007|2|28
1111119|2007|4|30
1111119|2007|12|31
1111119|2007|8|8

and let assume that today is june 30 2007 as an output I should have:

1111119|2007|2|28
1111119|2007|4|30
1111119|2007|12|31
Honored Contributor
Patrick Wallek
Posts: 13,784
Registered: ‎06-21-2000
Message 8 of 16 (327 Views)

Re: please help with kshell script..

I am thoroughtly confused as to what you are looking for as output.

Can you please state your EXACT requirements?
Regular Advisor
amonamon
Posts: 166
Registered: ‎05-21-2006
Message 9 of 16 (327 Views)

Re: please help with kshell script..

yes I know it is tricky..little..

but if today is for example june 30, 2007

and my input file is like this:

8111319|2007|2|28
1112119|2007|4|30
7111119|2007|12|31
1231119|2007|8|8

script should check if next day is 1st in the month and output this:

8111319|2007|2|28
1112119|2007|4|30
7111119|2007|12|31

becouse february has 28 days...if today is for example 8 sept or 8 dec or any other month output should be like this:

1231119|2007|8|8

that means if I have bigger file:

1234561|2007|2|28
4567891|2007|3|31
7897891|2007|4|5
9879871|2007|5|30
1578961|2007|6|28
1897893|2007|10|28
7879871|2007|12|30
9578962|2007|12|28

and today is 28 of august:

output should be like this:

1234561|2007|2|28
1578961|2007|6|28
1897893|2007|10|28
9578962|2007|12|28

that means does not metter which month is it looks fot day..


Honored Contributor
Patrick Wallek
Posts: 13,784
Registered: ‎06-21-2000
Message 10 of 16 (327 Views)

Re: please help with kshell script..

OK, so let me try to get this straight. Your requirements are:

1) If the current day is the LAST DAY of the month, then look in the file for ANY LINES from the current year that occur on the last day of their respective months.

2) If the current day is some other time in the month, then look for lines from the current year from any month where the day equals today.

Is that accurate?
Regular Advisor
amonamon
Posts: 166
Registered: ‎05-21-2006
Message 11 of 16 (327 Views)

Re: please help with kshell script..

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,

Honored Contributor
Peter Nikitka
Posts: 1,575
Registered: ‎02-10-2003
Message 12 of 16 (327 Views)

Re: please help with kshell script..

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
The Universe is a pretty big place, it's bigger than anything anyone has ever dreamed of before. So if it's just us, seems like an awful waste of space, right? Jodie Foster in "Contact"
Honored Contributor
Peter Nikitka
Posts: 1,575
Registered: ‎02-10-2003
Message 13 of 16 (327 Views)

Re: please help with kshell script..

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' ...

The Universe is a pretty big place, it's bigger than anything anyone has ever dreamed of before. So if it's just us, seems like an awful waste of space, right? Jodie Foster in "Contact"
Honored Contributor
john korterman
Posts: 1,117
Registered: ‎11-15-2000
Message 14 of 16 (327 Views)

Re: please help with kshell script..

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.
it would be nice if you always got a second chance
Regular Advisor
amonamon
Posts: 166
Registered: ‎05-21-2006
Message 15 of 16 (327 Views)

Re: please help with kshell script..

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

I got this:

grep: illegal option -- E
Usage: grep -hblcnsviw pattern file . . .
Honored Contributor
john korterman
Posts: 1,117
Registered: ‎11-15-2000
Message 16 of 16 (327 Views)

Re: please help with kshell script..

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.
it would be nice if you always got a second chance
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