Almost your entire problem is thinking that you have an awk script instead of a bash or ksh shell script. What you have is a shell pipeline where the first command in the pipeline happens to be awk. If your 1-liner works when typed into your shell from your terminal window, adding the full pathname of your shell after #! as the 1st line of your script and your 1-liner as the 2nd line of the script would work.

Note that the date utility is not a system call. I haven't seen any OS that has a date() system call. You may be thinking about the awk system() function which can be used to invoke a shell to run a command, but you don't want to invoke a shell from awk to process each line you're reading from a file. If you need to do that, you're probably wasting a lot of time by using awk instead of just reading the file directly in the shell.

Note that a 1-liner that long is impossible to read and update (assuming you may want to make changes later). So you should probably reformat your 1-liner into more readable chunks. You haven't said what shell you're using, so I'll use the shell I normally use (ksh) as an example. For this script ksh and bash are interchangeable:

Code:

#!/bin/ksh
awk -v RefDate="$(date -d'now-2 second' '+%s')" \
	-v LogDate="$(echo "$4 $5" |
		sed -e 's/\//-/g' -e 's/:/ /1' |
		tr -d '[]' |
		date '+%s')" '
{	if(LogDate > RefDate)
		print RefDate, LogDate, RefDate-LogDate, $0
}' "/var/log/httpd/rewrite.log" |
grep -B 1 -A 0 -inP '(rewrite\s+.?assets/.+.?\s+->\s+.?sites/default/files/.+.?|rewrite .?sites/default/files/.+.?\s+->\s+.?assets/.+.?)'

And, looking at this we can more easily see that all of the code marked in red above from your 1-liner is a no-op (the date utility does not read anything from its standard input; so the echo, sed, and tr commands don't do anything useful. And, the echo "$4 $5" is referring to the 4th and 5th arguments passed to your shell (not the 4th and 5th fields from a line in the file awk is reading, so that pipeline is just passing a <space> character followed by a <newline> character to date anyway).

So, this script can be further simplified to:

Code:

#!/bin/ksh
awk -v RefDate="$(date -d'now-2 second' '+%s')" \
	-v LogDate="$(date '+%s')" '
{	if(LogDate > RefDate)
		print RefDate, LogDate, RefDate-LogDate, $0
}' "/var/log/httpd/rewrite.log" |
grep -B 1 -A 0 -inP '(rewrite\s+.?assets/.+.?\s+->\s+.?sites/default/files/.+.?|rewrite .?sites/default/files/.+.?\s+->\s+.?assets/.+.?)'

and still produce the same output.
And, as long as RefDate is always set to now minus something and LogDate is always set to now, the script can be further simplified to:

Code:

#!/bin/ksh
awk -v RefDate="$(date -d'now-2 second' '+%s')" \
	-v LogDate="$(date '+%s')" '
{	print RefDate, LogDate, RefDate-LogDate, $0
}' "/var/log/httpd/rewrite.log" |
grep -B 1 -A 0 -inP '(rewrite\s+.?assets/.+.?\s+->\s+.?sites/default/files/.+.?|rewrite .?sites/default/files/.+.?\s+->\s+.?assets/.+.?)'

and still produce the same output. Note also that since the input line you showed us does not contain the string "rewrite", grep is not going to match any lines and you won't have any output. I, therefore, have to assume that there are other lines in your log file that are in a different format.

If you will show us a more representative sample of your input file, and be more clear about what you're really trying to do, I think I can help you reformat the way you set RefDate so we can compare that to the date and time in the 4th and 5th fields from lines in the format you showed us. But, unless the lines you're using to match with the grep have the same data in the 4th and 5th fields, you aren't going to get the output you want.

PLEASE show us some representative sample input AND show us the output you want to get from your script!

If you have the dates and times shown in human readable terms (instead of a seconds since the Epoch), does your output really need the number of seconds between the two? If you need that difference to be shown in seconds, your script will run MUCH slower.