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awk - problems by converting date-format

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# 1  
Old 01-09-2014
awk - problems by converting date-format
Hi

i try to change the date-format from DD/MM/YYYY into MM/DD/YY.
Input-Data:
Code:
...
31/12/2013,23:40,198.00,6.20,2,2,2,1,11580.0,222
31/12/2013,23:50,209.00,7.30,2,2,3,0,4380.0
01/01/2014,00:00,205.90,8.30,2,2,3,1,9360.0,223
...

Output-Data should be:
Code:
...
12/31/13,23:40,198.00,6.20,2,2,2,1,11580.0,222
12/31/13,23:50,209.00,7.30,2,2,3,0,4380.0
01/01/14,00:00,205.90,8.30,2,2,3,1,9360.0,223
...

i try to do it with awk ....
Code:
awk -F'[,|/]' '{ printf "%2d/%02d/%02d,%s,%s,%s\n" $2,$1,$3,$NF}'

but awk "ran out" -- i have still not understand, what awk do with the tripple %s. can someone help me please.

Thanks in advance,
IMPe
# 2  
Old 01-09-2014
"%2d/%02d/%02d,%s,%s,%s\n" is expecting 6 arguments, and you only have 4 (and 11 or 12 fields on the lines).

One way might be to use split rather than having multiple field separators:
Code:
awk '{split($1,d,"/"); $1=d[2]"/"d[1]"/"d[3]; print}' FS=OFS=, file

This User Gave Thanks to CarloM For This Post:
IMPe
# 3  
Old 01-09-2014
Quote:
Originally Posted by CarloM
"%2d/%02d/%02d,%s,%s,%s\n" is expecting 6 arguments, and you only have 4 (and 11 or 12 fields on the lines).

One way might be to use split rather than having multiple field separators:
Code:
awk '{split($1,d,"/"); $1=d[2]"/"d[1]"/"d[3]; print}' FS=OFS=, file

Hi CarloM,

Thanks a lot for your fast reply. Your "split-version" is working pretty good, but the the Year has still four numbers.
It is possible, to combine it with something like substr(d[3],3,2) ?
Code:
awk '{split($1,d,"/"); $1=d[2]"/"d[1]"/"(substr(d[3],3,2)); print}' FS=OFS=, file

only give me the formated date in MM/DD/YY.
I'm still try to become more familliar with awk and i suppose split will help me several times more in the future.

Thanks,
IMPe
Last edited by IMPe; 01-09-2014 at 09:44 AM..
# 4  
Old 01-09-2014
I suspect d[3] may actually be the rest of the line. Try setting FS & OFS separately.
Code:
awk '{split($1,d,"/"); $1=d[2]"/"d[1]"/"(substr(d[3],3,2)); print}' FS=, OFS=, file

This User Gave Thanks to CarloM For This Post:
IMPe
# 5  
Old 01-09-2014
Could be done with simple sed statements like:
Code:
sed 's!\(..\)/\(..\)!\2/\1!' yourfile

or
Code:
sed 's!\(...\)\(...\)!\2\1!' yourfile

This User Gave Thanks to ctsgnb For This Post:
IMPe
# 6  
Old 01-09-2014
CarloM and ctsgnb, to both of you a big thank you.
Quote:
Originally Posted by ctsgnb
Could be done with simple sed statements like:
Code:
sed 's!\(..\)/\(..\)!\2/\1!' yourfile

or
Code:
sed 's!\(...\)\(...\)!\2\1!' yourfile

ctsgnb --> this sed-command will change the "position" of dd/mm to mm/dd but it will not reduce YYYY to YY. finally i want to compare the time-stamp with data from another file, so i need the year-format absolutly like YY.

Thanks,
IMPe
# 7  
Old 01-09-2014
Code:
 sed 's#\(..\)/\(..\)/..\(..\)#\2/\1/\3#' file
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