Print line containing pattern preceded by different line containing a different pattern

macOS 13.3 Ventura hence BSD versions of grep, awk, et al.

How do I search for and print a line containing a pattern where the line MUST be preceded by a different line containing a different pattern?

The text contains lines like these (leading CAPS for reference, ...==irrelevant chars).

  ...                   # An indeterminate number of lines.
A ... "model" = \< ...
  ...                   # An indeterminate number of lines.
B ... PXS4@0 ...
  ...                   # An indeterminate number of lines.
C ... "model" = \< ...
  ...                   # An indeterminate number of lines.
D ... PXS2@0 ...
  ...                   # An indeterminate number of lines.
E ... "model" = \< ...
  ...                   # An indeterminate number of lines.
F ... PXS1@0 ...
  ...                   # An indeterminate number of lines.
G ... "model" = \< ...
  ...                   # An indeterminate number of lines.
H ... "model" = \< ...
  ...                   # An indeterminate number of lines.

ONLY lines with "model" that are preceded by a line with PXS[[:digit:]]@0 should appear:

C ... "model" = \< ...
E ... "model" = \< ...
G ... "model" = \< ...

AFAICT the regex in macOS's awk & grep do not support look-behind and look-ahead.

I thought this would find a match of PXS... and then find/print model... but it prints line "A":

awk '/(PXS\[\[:digit:\]\]@0 )+?model" = \</ { print }'

This also comes close but prints line "A". Since it prints "A" I don't understand why it doesn't also print "H".

grep -e ".\*PXS\[\[:digit:\]\]@0 " -e ".\*model" = \<"" | grep -v -e ".\*PXS\[\[:digit:\]\]@0 "

Enlighten me please!

MacOs

perl -n0e 'print $_ =~ /PXS[[:digit:]]@0.*\n.*\n/g' filename | perl -p -e 's/PXS[[:digit:]]@0[^\n]*\n//g'

First step: leave only lines with PXS[[:digit:]]@0 and the next lines.
Second step: remove lines with PXS[[:digit:]]@0

Linux

grep -zoP 'PXS[[:digit:]]@0.*\n.*\n' filename | sed -z -E 's/PXS[[:digit:]]@0[^\n]*\n//g'

Grep to find lines with PXS[[:digit:]]@0 and the next lines, sed to remove lines containing PXS[[:digit:]]@0 from output.

you can try something like

awk '/PXS[0-9][@]0/{getline;if(match($0,"model")){ print;}}' 

/PXS[0-9][@]0/ will match the prefix line

getline; will read the next line (and populate $0)

match($0,"model") will see if that line contains the regexp 'model'

Thanks for steering me in the right direction. This gives me lines C, E, and G.

I used a loop in awk to find the first line with PXS[[:digit:]]@0 and a sub loop to find the second with "model" = <. The file is deterministic: if the first line is present, the second will be (but not directly after the first).

I also set awk's delimiters since the final value I want is after "model" = \<" and before ">.

awk -F'<"|">' 'BEGIN {while (getline != 0) if ($0 ~ /PXS[[:digit:]]@0 /) {while (getline != 0) if ($0 ~ /"model" = </){print $2; break;}}}'

I like having the whole thing in one awk command, my other solution required 5 piped greps.

$ awk '/model/ && match(prevline,/PXS[0-9]@0/){print} {prevline=$0}' file
C ... "model" = \< ...
E ... "model" = \< ...
G ... "model" = \< ...