unix time to date and replace in bash using awk -

i trying convert unix time date , time;

1436876820 blah1 stop none john 1436876820 blah0 continu none john 1436876821 blah2 stop bob 

i convert first column have 2 more column date , time below

14-07-15 13:27:00 blah1 stop none john 14-07-15 13:27:00 blah0 continu none john 14-07-15 13:27:01 blah2 stop bob etc.. 

so have started following.

in="${1}"  in $(awk '{print $1}' ${in});  dd=$(date -d @${i} +'%d-%m-%y %h:%m:%s')  awk '{ ${1}="'"${dd}"'" }' < ${in}   done 

this not work due syntax , give such of error:

awk: { ${1}="14-07-2015 13:27" } awk:          ^ syntax error 

i use sed instead of awk:

sed "s/^1........./${dd}/" ${in} 

any awk welcome. al.

get rid of shell loop , 1 awk invocation:

awk '{     cmd = "date -d @" $1 " +\"%d-%m-%y %h:%m:%s\""     if ( (cmd | getline dd) > 0 ) {         $1 = dd     }     close(cmd)     print }' "$1" 

if have gnu awk can use it's internal strftime() instead of date+getline:

awk '{     $1 = strftime("%d-%m-%y %h:%m:%s",$1)     print }' "$1"