Print all the lines between two patterns in shell -

i have file log of script running in daily cronjob. log file looks like-

aug 19  line1 line2 line3 line4 line5 line6 line7 line8 line9  aug 19  aug 20  line1 line2 line3 line4 line5 line6 line7 line8 line9  aug 20  aug 21  line1 line2 line3 line4 line5 line6 line7 line8 line9  aug 21 

the log written script starting date , ending date , in between logs written.

now when try logs single day using command below -

sed -n '/aug 19/,/aug 19/p' filename 

it displays output -

aug 19  line1 line2 line3 line4 line5 line6 line7 line8 line9  aug 19 

but if try logs of multiple dates, logs of last day missing.

example- if run command

sed -n '/aug 19/,/aug 20/p' filename 

the output looks -

aug 19  line1 line2 line3 line4 line5 line6 line7 line8 line9  aug 19  aug 20 

i have gone through site , found valuable inputs similar problem none of solutions work me. links link 1

link 2

the commands have tried -

awk '/aug 15/{a=1}/aug 21/{print;a=0}a' awk '/aug 15/,/aug 21/' sed -n '/aug 15/,/aug 21/p grep -pzo "(?s)(aug 15(.*?)(aug 21|\z))" 

but none of commands gives logs of last date, commands prints till 1st timestamp have shown above.

i think can use awk command followed print lines between aug 19 & aug 20,

awk '/aug 19/||/aug 20/{a++}a; a==4{a=0}' file 

brief explanation,

• /aug 19/||/aug 20/: find record matched aug 19 or aug 20
• if criteria met, set flag a++
• if flag a in front of semicolon greater 0, print record.
• final criteria, if a==4, reset a=0, mind worked case in example, if aug 19 or aug 20 more 4, modify number 4 in answer meet new request.

if want assign searched patterns variables, modify command followed,

$ b="aug 19" $ c="aug 20" $ awk -v b="$b" -v c="$c" '$0 ~ c||$0 ~ b{a++}a; a==4{a=0}' file 

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