bash - regex backward find all text between a string and first occurrence of another string -

i need find recent occurrence of 'get' ( zzzz) before 'error:' , capture of text in between.

get xxxxx yyyyy zzzzz text more text error: error 

can done?

edit

thanks, awk solution works, can further improved getting last occurrence of 'error:' ?

get xxxxx yyyyy zzzzz text more text error: first error  xxxxx yyyyy zzzzz text more text error: last error 

try following awk solution:

awk '   /^get/ { delete lines; c=0; inblock=1 }   /^error:/ { for(i=1; i<=c; ++i) print lines[i]; print; exit }   inblock { lines[++c] = $0 } ' file 

this assumes 1 block must printed, , error: line should printed. (update: see below solution prints last block).

• /^get/ { delete lines; c=0; inblock=1 } starts building array of lines in variable lines whenever string get encountered @ start of line.
• /^error:/ { for(i=1; i<=c; ++i) print lines[i]; print; exit } matches string error: @ start of line , prints out lines built far, followed current line, , exits.
• inblock { lines[++c] = $0 } adds every line starting recent get line array.

---

update, per op's request:

to report (only) last block ends error:, use following:

awk '   /^get/ { delete lines; c=0; inblock=1 }   inblock { lines[++c] = $0 }   /^error:/ { inblock=0; }   end { for(i=1; i<=c; ++i) print lines[i] } ' file 

this differs first solution in later blocks replace earlier ones, last block "wins", printed after input has been processed, in end block of awk script.