Changing date format

Shell Programming and Scripting
1

Hi, I have a column in a table of Timestamp datatype. For Example : Var1 is the column 2008-06-26-10.10.30.2006. I have Given query as date(var1) and time (var1) I got the file as in the below format :

File1:

Col1 Col2
2008-06-02|12.36.06
2008-06-01|23.36.35

But the problem is I need the output as

Col1 Col2
02062008|12:36:06
01022008|23:36:35

Thanks in advance.

2

`date +%Y%m%d`
will give ya date in YYYYMMDD format

`date +%H:%M:%S`

will give ya time in HH:MM:SS format

Hope this helps

3

awk can do the job:

$ cat File1
2008-06-02|12.36.06
2008-06-01|23.36.35
$ awk -F'[-|.]' '{printf "%s%s%s%c%s%c%s%c%s\n",$3,$2,$1,124,$4,58,$5,58,$6}' File1
02062008|12:36:06
01062008|23:36:35
4 
echo '2008-06-02|12.36.06' | nawk -F'([|]|[-])' '{gsub("[.]", ":", $4); print $3$2$1 "|" $4}'
5

Thanks Danmero and vgresh99 for the reply. But still I have some another problem. If I have my file as

File1:

Col1 Col2 ID
2008-06-02|12.36.06|1-432
2008-06-01|23.36.35|1-345

In the above file I have ID column values as 1<<Hypen>>432,1<<hypen>>345.

What's happening is when I give the command
$ cat File1
$ nawk -F'([|]|[-])' '{gsub("[.]", ":", $4); print $3$2$1 "|" $4 "|" $5

It is giving output as :

Col1 Col2 ID
2008-06-02|12.36.06|1|432
2008-06-01|23.36.35|1|345

It is replacing 1<<Hypen>>432 as 1|432. Please let me know how to get out of it. Thanks in advance.

6

Below code is more flexible and can accommodate for more fields in the data input:

$ cat b.sh
#! /bin/sh

if [ $# -ne 1 ]; then
        echo "Usage: $0 <input.txt>"
        exit 1
fi


awk '
BEGIN {
        FS="|"
}
{
        split($1,ymd,"-")
        split($2,hms,".")
        d=sprintf("%s%s%s",ymd[3],ymd[2],ymd[1])
        t=sprintf("%s:%s:%s",hms[1],hms[2],hms[3])

        printf("%s|%s",d,t)
        for(i=3;i<=NF;++i) {
                printf("|%s",$i)
        }
        printf("\n")
}' $1

$ cat b.in
2008-06-02|12.36.06|32-143|1234.23
2008-06-01|23.36.35|12-239|99.45

$ ./b.sh b.in
02062008|12:36:06|32-143|1234.23
01062008|23:36:35|12-239|99.45
7 
#!/usr/bin/ksh93

IFS='|'
while read col1 col2 col3
do
    printf "%(%d%m%Y|%T)T|%s\n" "$col1 $col2" $col3
done < file

$./t
02062008|12:36:06|1-432
01062008|23:36:35|1-345
$
8

I think Chihung's input will be good. I tried it before. It worked for me.