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date(ddmmyyyy) sorting

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# 1  
Old 12-06-2011
date(ddmmyyyy) sorting
input :
Code:
20110730
20110730
20110731
20110731
20110801
20110801
20110801
20110813
20110815
01062011
01062011

OUTPUT : i need to sort this input in such a way so that the latest date comes first.
Last edited by radoulov; 12-06-2011 at 10:50 AM.. Reason: Code tags!
# 2  
Old 12-06-2011
And what have you tried so far?
Last edited by radoulov; 12-06-2011 at 11:01 AM..
# 3  
Old 12-06-2011
If you reorder that into YYYYMMDD then an ordinary alphabetic sort will get it.
# 4  
Old 12-07-2011
input.txt:
Code:
20110730
20110730
20110731
20110731
20110801
20110801
20110801
20110813
20110815
01062011
01062011

script.sh:
Code:
#! /bin/bash
while read x
do
    echo $x | egrep -q "20[0-9][0-9](0[1-9]|1[0-2])(0[1-9]|[12][0-9]|3[01])"
    if [ $? -eq 0 ]
    then
        y=`echo $x | cut -c7-8``echo $x | cut -c5-6``echo $x | cut -c1-4`
        sed -i "s/$x/$y/" input.txt
    fi
done < input.txt
sort -n -k1.5 -k1.3 -k1 input.txt

Code:
$ ./script.sh
01062011
01062011
30072011
30072011
31072011
31072011
01082011
01082011
01082011
13082011
15082011

Last edited by balajesuri; 12-07-2011 at 01:06 AM..
# 5  
Old 12-07-2011
Mileage may vary:

Code:
echo "12319999\n01012013\n12312011\n01012011" |sort -k1.4 -k5

# 6  
Old 12-07-2011
@curleb: In the input file as given by urfrnddpk, the last line is "01062011", how would one know if its DDMMYYYY or MMDDYYYY? I've assumed it as DDMMYYYY.
# 7  
Old 12-07-2011
Quote:
Originally Posted by curleb
Mileage may vary:

Code:
echo "12319999\n01012013\n12312011\n01012011" |sort -k1.4 -k5


thanks for the solution
can you also tell me the logic of parameters you have given (-k1.4 -k5)??
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