How can I make grep print the lines below and above each matching line

Question

I have to parse a very large file and I want to use the command grep  or any other tool    I want to search each log line for the word FAILED  then print the line above and below each matching line  as well as the matching line   For example   id   15 Satus   SUCCESS Message   no problem     id   15 Satus   FAILED Message   connection error     And I need to print   id   15 Satus   FAILED Message   connection error

User · Accepted Answer

grep s -A 1 option will give you one line after  -B 1 will give you one line before  and -C 1 combines both to give you one line both before and after  -1 does the same

User · Answer

Use -A and -B switches  mean lines-after and lines-before    grep -A 1 -B 1 FAILED file txt

User · Answer

Use -B  -A or -C option  grep --help     -B  --before-context NUM  print NUM lines of leading context -A  --after-context NUM   print NUM lines of trailing context -C  --context NUM         print NUM lines of output context -NUM                      same as --context NUM

[linux] How can I make grep print the lines below and above each matching line?

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