CODEWITHSUNDEEP
bashshellunixawk
Akshay Singh
Akshay Singh

Reputation: 127

Using awk how to fetch the date day wise?

I'm stuck how to fetch the date in day wise i.e if date is like 03/12/2020 how can I fetch only "03" in awk? Let's make a scenario using dummy data, file records are given something as follows

Cust_id;  orderDate; orderAmt
101; 20/09/2020; 8000
102; 08/08/2020; 6000
103; 03/11/2020; 12000
104; 21-06-2020; 1200
105; 04-07-2020; 1300

Possible Output:

103; 12000
105; 1300

how can I print the details of customer who bought item in very first week in any month using AWK? How to fetch the date field with day wise here. Also note that if record like 104 'id', if date format is changed i.e. 21-06-2020 from default 21/06/2020 how to deal it?

I have tried on stack overflow but didn't get this date situation problem and also tried below link couldn't help me more ?

https://www.unix.com/shell-programming-and-scripting/138974-using-awk-print-date.html

Upvotes: 1

Views: 616

Answers (3)

stack0114106
stack0114106

Reputation: 8711

Another solution:

I assume the OP is referring to purchases within first 7 days of a month. Form the given input date it looks like the first part is the day, 2nd one is month. Multiple delimiters can be applied and the second one can be compared with < 7 days.

$ cat akshay.txt
Cust_id;  orderDate; orderAmt
101; 20/09/2020; 8000
102; 08/08/2020; 6000
103; 03/11/2020; 12000
104; 21-06-2020; 1200
105; 04-07-2020; 1300
$ awk -F"[-;/ ]+" ' $2<7 ' akshay.txt
103; 03/11/2020; 12000
105; 04-07-2020; 1300
$

Upvotes: 1

RavinderSingh13
RavinderSingh13

Reputation: 133458

Adding one more variant here, could you please try following written and tested with shown samples. Pass date in DDMMYYYY format in awk's variable named dat with any dash or slashes and inside it will compare with 2nd field's substring(s) of it.

awk -v dat="04072020" '
BEGIN{
  FS="[[:space:]]*;[[:space:]]*"
  OFS="; "
}
dat == (substr($2,1,2) substr($2,4,2) substr($2,7,4)){
  print $1,$3
}
' Input_file

Explanation: Adding detailed explanation for above.

awk -v dat="04072020" '              ##Starting awk program from here and creating dat with date of DDMMYYYY format.
BEGIN{                               ##Starting BEGIN section of this program from here.
  FS="[[:space:]]*;[[:space:]]*"     ##Setting field separator spaces(one or more occurrences) semi colon and again one or more occurrences of spaces.
  OFS="; "                           ##Setting output field separator as semi colon space here.
}
dat == (substr($2,1,2) substr($2,4,2) substr($2,7,4)){ 
                                     ##Checking condition if dat is equal to substring of (1st,2nd)chars (4th,5th) and (7th,8th) chars collective sub string.
  print $1,$3                        ##Printing 1st and 3rd fields here.
}
' Input_file                         ##Mentioning Input_file name here.

Upvotes: 2

Ed Morton
Ed Morton

Reputation: 203334

Is this what you're trying to do?

$ cat tst.awk
BEGIN { FS="[[:space:]]*;[[:space:]]*"; OFS="; " }
NR == 1 { next }
{ gsub("-","/",$2) }
$2 == tgt { print $1, $3 }


$ awk -v tgt='03/11/2020' -f tst.awk file
103; 12000

$ awk -v tgt='04/07/2020' -f tst.awk file
105; 1300

Upvotes: 3

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