Shell script to parse incoming dates and print in

问题:

I need a shells script which could parse incoming date values and print this in a standard format.

Incoming date patterns are:

    "yyyyMMdd HHmmss"
    "yyyyMMdd_HHmmss"
    "MMddyyyy:HHmmss"
    "MMddyyyyHHmmssmillisecond"

     20170426 102300
     20170426_102300
     04262017:102300
     0426201710230066

Output date pattern :

yyyyMMdd_HHmmSS 20170426_102300

Any idea how to achieve this result in bash. I tried couple of regex for getting result , but that didn't help. Any help is appreciated.

回答1:

Pipe the input to sed:

sed -re 's/([0-9]{8}) ([0-9]{6})/\1_\2/' -e 's/([0-9]{4})([0-9]{4}):?([0-9]{6}).*/\2\1_\3/'

回答2:

Is perl acceptable?

while (<>) {
    if ($_ =~ m/(\d{8})[ _](\d{6})/) {
        print "$1_$2";
    } elsif ($_ =~ m/(\d{4})(\d{4}):(\d{6})/) {
        print "$2$1_$3";
    } elsif ($_ =~ m/(\d{4})(\d{4})(\d{6})\d*/) {
        print "$2$1_$3";
    }
}

Havn't tested it though...

You could also use it like this:

~$ cat data | perl -e 'while (<>) {
    if ($_ =~ m/(\d{8})[ _](\d{6})/) {
        print "$1_$2";
    } elsif ($_ =~ m/(\d{4})(\d{4}):(\d{6})/) {
        print "$2$1_$3";
    } elsif ($_ =~ m/(\d{4})(\d{4})(\d{6})\d*/) {
        print "$2$1_$3";
    }
}'

For your array this may be acceptable:

~$ perl -e 'for (@ARGV) {
    if ($_ =~ m/(\d{8})[ _](\d{6})/) {
        print "$1_$2\n";
    } elsif ($_ =~ m/(\d{4})(\d{4}):(\d{6})/) {
        print "$2$1_$3\n";
    } elsif ($_ =~ m/(\d{4})(\d{4})(\d{6})\d*/) {
        print "$2$1_$3\n";
    } else { print "$_ does not fit\n"; }
}' "${testdata[@]}"

If you don't have perl on your production-environment, you probably want to settle for a sed solution. I suggest the one from Walter A:

for t in "${testdata[@]}"; do 
    echo $t |  sed -re 's/([0-9]{4})([0-9]{4})([0-9:])/\2\1\3/; s/[ _:]//;s/(.{8})(.{6}).*/\1_\2/'; 
done

回答3:

For fun, here is a solution using awk:

awk 'NF==2{print $1"_"$2} $1~"_"{print $1} $1~":"{print gensub(/([0-9]{4})([0-9]{4}):([0-9]{6})/, "\\2\\1_\\3", "g", $1)} length($1)==16{print gensub(/([0-9]{4})([0-9]{4})([0-9]{6}).*/, "\\2\\1_\\3", "g", $1)}'

Pretty much the same as the perl and sed examples. Testing and regex replacing.

回答4:

First make all dates in the format yyyyMMdd

sed -r 's/([0-9]{4})([0-9]{4})([0-9:])/\2\1\3/'

Next remove the optional character between day and hour

sed -r 's/([0-9]{4})([0-9]{4})([0-9:])/\2\1\3/; s/[ _:]//'

Change yyyyMMddHHmmss?? into desired format

sed -r 's/([0-9]{4})([0-9]{4})([0-9:])/\2\1\3/; s/[ _:]//;s/(.{8})(.{6}).*/\1_\2/' 

EDIT: I first tried to show msec, but those were not needed:

# INCORRECT SOLUTION
# sed -r 's/([0-9]{4})([0-9]{4})([0-9:])/\2\1\3/; s/[ _:]//; s/$/00/' | cut -c1-16