Replace a date string in sed with the ouput of a custom function with arguments

Question

I have a bash script which echos out an html file like this ./foo.sh > out.html In the html there are timestamps in the following format 2019-02-24T17:02:19Z.

I wrote a function to convert the time stamp to the time delta between the timestamp and now.

my_uptime(){
    START=$(date -d "$1" "+%s")
    NOW=$(date "+%s")
    STD=$(echo "($NOW-$START)/3600" | bc)
    MIN=$(echo "(($NOW-$START)/60)%60" | bc)
    echo "Uptime $STD h $MIN min"
}

Now I want to replace the timestamp with the output of my_uptime directly in the stream. I tried this:

echo "<p>some html</p>
2019-02-24T17:02:19Z
<p>some more html</p>" | sed -r "s/[0-9\-]+T[0-9:]+Z/$(my_uptime \0)/"

This fails because the command substitution doesn't recognize the back reference and puts in a literal 0. Is there another way to achieve this? Preferably directly in the stream.

Answer 1

... | sed -r "s/[0-9\-]+T[0-9:]+Z/$(my_uptime \0)/"

This code is attempting to pass the matched value from sed's s/// into the shell function. However, $(...) is expanded before sed even sees it.

Using sed is probably not appropriate here.

Here's a perl replacement that effectively combines your shell function and sed:

... | perl -ple '
    if (/([0-9-]+T[0-9:]+Z)/) {
        $s = `date -d "$1" +%s`;
        $n = time;
        $h = int(($n-$s)/3600);
        $m = int(($n-$s)/60)%60;
        $_ = "Uptime $h h $m min";
    }'

You could probably do something similar in awk.