Substitute value with result of calling function o

问题:

I have a text stream that looks like this:

----------------------------------------
s123456789_9780
  heartbeat:test       @ 1344280205000000: '0'
  heartbeat:test       @ 1344272490000000: '0'

Those long numbers are timestamps in microseconds. I would like to run this output through some sort of pipe that will change those timestamps to a more human-understandable date.

I have a date command that can do that, given just the timestamp (with the following colon):

$ date --date=@$(echo 1344272490000000: | sed 's/.......$//') +%Y/%d/%m-%H:%M:%S
2012/06/08-10:01:30

I would like to end up with something like this:

----------------------------------------
s123456789_9780
  heartbeat:test       @ 2012/06/08-12:10:05: '0'
  heartbeat:test       @ 2012/06/08-10:01:30: '0'

I don't think sed will allow me to match the timestamp and replace it with the value of calling a shell function on it (although I'd love to be shown wrong). Perhaps awk can do it? I'm not very familiar with awk.

The other part that seems tricky to me is letting the lines that don't match through without modification.

I could of course write a Python program that would do this, but I'd rather keep this in shell if possible (this is generated inside a shell script, and I'd rather not have dependencies on outside files).

回答1:

Bash with a little sed, preserving the whitespace of the input:

while read -r; do                                                                                                                                                                                                                                          
    parts=($REPLY)
    if [[ ${parts[0]} == "heartbeat:test" ]]; then
        dateStr=$(date --date=@${parts[2]%000000:} +%Y/%d/%m-%H:%M:%S)
        REPLY=$(echo "$REPLY" | sed "s#[0-9]\+000000:#$dateStr#")
    fi
    printf "%s\n" "$REPLY"
done

回答2:

This might work for you (GNU sed):

sed '/@ /!b;s//&\n/;h;s/.*\n//;s#\(.\{10\}\)[^:]*\(:.*\)#date --date=@\1 +%Y/%d/%m-%H:%M:%S"\2"#e;H;g;s/\n.*\n//' file

Explanation:

• /@ /!b bail out and just print any lines that don't contain an @ followed by a space
• s//&\n/ insert a newline after the above pattern
• h copy the pattern space (PS) to the hold space (HS)
• s/.*\n// delete upto and including the @ followed by a space
• s#\(.\{10\}\)[^:]*\(:.*\)#date --date=@\1 +%Y/%d/%m-%H:%M:%S"\2"#e from whats remaining in the PS, make a back reference of the first 10 characters and from the : to the end of the string. Have these passed in to the date command and evaluate the result into the PS
• H append the PS to the HS inserting a newline at the same time
• g copy the HS into the PS
• s/\n.*\n// remove the original section of the string

回答3:

How about:

while read s1 at tm s2
do 
    tm=${tm%000000:}
    echo $s1 $at $(date --date @$tm +%Y/%d/%m-%H:%M:%S)
done < yourfile

回答4:

I would also like to see a sed solution, but it is a bit beyond my sed-fu. As awk supports strftime it is fairly straight forward here:

awk '
/^ *heartbeat/ { 
  gsub(".{7}$", "", $3)
  $3 = strftime("%Y/%d/%m-%T", $3)
  print " ", $1, $3
}

$0 !~ /heartbeat/' file

Output:

s123456789_9780
heartbeat:test 2012/06/08-21:10:05
heartbeat:test 2012/06/08-19:01:30

$3 is the microsecond field. gsub converts the timestamp to seconds.

The $0 !~ makes sure non-heartbeat lines are printed ({ print } implicitly is the default block).

回答5:

This does it mostly within bash using your date command:

#!/bin/bash
IFS=$
while read a ; do
case "$a" in
*" @ "[0-9]*) pre=${a% @ *}
              a=${a#$pre @ }
              post=${a##*:}
              a=${a%??????:$post}
              echo "$pre$(date --date=@$a +%Y/%d/%m-%H:%M:%S):$post"
              ;;
*)            echo "$a" ;;
esac
done <<.
----------------------------------------
s123456789_9780
  heartbeat:test       @ 1344280205000000: '0'
  heartbeat:test       @ 1344272490000000: '0'
.