使用包含字段文本的命令在 awk 中获取 shell 命令的 output

Question

I'm trying to parse a file for some dates in a specific format and get the 'seconds since the Epoch' to calculate absolute differences between the date and the current date. I need to do this on each line using the 4th field, so I tried doing something like this:

"date --date=$4 +%s" | getline DUE
close("date --date=$4 +%s")
"date +%s" | getline CURR
close("date +%s")

Of course this doesn't quite work because awk interprets the $4 as a literal "$4". So I can't use variables inside that statement.

I also tried creating 2 variables with the front of the command string and the ending with a third variable being those concatinated with $4 in the middle. That didn't work as well.

I have a feeling you should be able to do such a thing. I know date parses the date just fine. The format is something like 'Thu Aug 11 11:40:00 UTC 2023'.

Is there any way to do this in awk? I've gone through a lot of documentation without ever seeing an example of fields being processed in a shell command.

Answer 1

Use string concatenation.

You also need quotes around the --date argument, since it contains whitespce.'

command = "date --date=\"" $4 "\" +%s"
command | getline DUE
close(command)
command2 = "date +%s"
command2 | getline CURR
close(command2)

Answer 2

side note about date utility::: the syntax is really different between gnu/linux one and BSD ones:

• gnu one allows for fractional epochs for input and output
• the @ is a mandatory requirement for epochs

gdate -d '@16602518201915245' # GNU-date (GNU coreutils 9.1)

Mon Nov  8 22:07:25 EST 526114827

• BSD one won't accept @ or non-integers

 date -r 16602518201915245 # BSD-date (from FreeBSD 12.0)

Mon Nov  8 22:07:25 EST 526114827

they do share one common feature though:

• an upper-limit for the year -1 + 2 ^ ( 3 ^ 3 + 4 )