Date Format

biju.mp · #1 (**permalink**) 12-31-2007

User entering Date in the below format.

Input date : 2007 12 31

Output expecting : Mon Dec 31

how can I convert in this format?

Thanks & Regards,
Biju MP

ghostdog74 · #2 (**permalink**) 12-31-2007

Code:

#!/bin/sh
s="2007 12 31"
year=`echo $s | cut -d" " -f1`
mon=`echo $s | cut -d" " -f2`
day=`echo $s | cut -d" " -f3`
cal $2 $1 | awk 'BEGIN{ 
                         d["Su"]="Sun"
                         d["Mo"]="Mon"
                         d["Tu"]="Tue"
                         d["We"]="Wed"
                         d["Th"]="Thu"
                         d["Fr"]="Fri"
                         d["Sa"]="Sat"
                         y[1]="Su"
                         y[2]="Mo"
                         y[3]="Tu"
                         y[4]="We"
                         y[5]="Th"
                         y[6]="Fr"
                         y[7]="Sa"
                         month["01"]="Jan"
                         month["02"]="Feb"
                         month["03"]="Mar"
                         month["04"]="Apr"
                         month["05"]="May"
                         month["06"]="Jun"
                         month["07"]="Jul"
                         month["08"]="Aug"
                         month["09"]="Sep"
                         month["10"]="Oct"
                         month["11"]="Nov"
                         month["12"]="Dec"
}

NR>2{
    for(i=1;i<=NF;i++){
        if ( $i == '"$day"' ){
            print d[y[i]],month['"$mon"'],$i
        }
    
    }
}'

output:

Code:

# ./test.sh
Mon Dec 31

fpmurphy · #3 (**permalink**) 12-31-2007

If the version of date on your OS supports the -d option and portability is not a major requirement, you could do something like

Code:

date -d "2007-12-31" "+%a %b %d"

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