please help with kshell script..

[amonamon]

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...

[Peter Godron]

Hi,
how about:

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

[Peter Godron]

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

[john korterman]

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.

[SANTOSH S. MHASKAR]

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

[SANTOSH S. MHASKAR]

Hi,

Sorry for a small mistake,

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

-Santosh

[amonamon]

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

[Patrick Wallek]

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

Can you please state your EXACT requirements?

[amonamon]

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..

[Patrick Wallek]

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?

[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.