CODEWITHSUNDEEP
bashawksed
Prashant
Prashant

Reputation: 21

bash search text between first occurence of pattern1 and last occurence of pattern2

I am trying to get all lines between first occurrence of pattern1 and last occurrence of pattern 2 both the patterns are regex

Example code

TEXT
TEXT
[SUN_START]
[SUN_END]

[MON_START]
TEXT
[MON_END]

[TUE_START]
[TUE_END]

[WED_START]
TEXT
[WED_END]
TEXT
TEXT

Output that I am expecting is

[SUN_START]
[SUN_END]

[MON_START]
TEXT
[MON_END]

[TUE_START]
[TUE_END]

[WED_START]
TEXT
[WED_END]

Pattern is XXX_START and XXX_END

What I am got so far is

cat /u01/app/oracle/admin/LNOPP1P/config/dbbackup_LNOPP1P.config | sed -n -e '/[[A-Z][A-Z][A-Z]_START]/,/[[A-Z][A-Z][A-Z]_END]/p'

But this does not keep the line breaks and displays everything together like this

[SUN_START]
[SUN_END]
[MON_START]
TEXT
[MON_END]
[TUE_START]
[TUE_END]
[WED_START]
TEXT
[WED_END]

I also want to make sure that it only matches the line starts with [[A-Z]_START] and same for END

Upvotes: 1

Views: 128

Answers (4)

anishsane
anishsane

Reputation: 20980

This awk should work:

awk '/_START\]/{p=1} p{a = a $0 ORS}/_END\]/{printf "%s", a; a="";}' file

Simple logic:

  1. At the first *_START tag, enable p=1. This will discard those lines before the first *_START tag.
  2. For every line, append the current line to a local variable.
  3. At every *_END tag, print the local variable and empty it.
  4. Since we are printing only at the *_END tag, those lines after the last *_END are not printed.

Upvotes: 1

Ed Morton
Ed Morton

Reputation: 203532

IMHO a two-pass approach without saving the contents in memory is the simplest and most robust:

$ awk '
    NR==FNR { if (/\[[A-Z]+_START\]/ && !beg) beg=NR; if (/\[[A-Z]+_END\]/) end=NR; next }
    FNR>=beg && FNR<=end
' file file
[SUN_START]
[SUN_END]

[MON_START]
TEXT
[MON_END]

[TUE_START]
[TUE_END]

[WED_START]
TEXT
[WED_END]

Consider using [[:upper:]] instead of [A-Z] for portability across locales.

I just saw you had this comment under a different answer: 

Is it simple to invert this selection? select everything but the bit selected by this AWK ?

and the answer is "of course", just change the condition at the end of the script:

$ awk '
    NR==FNR { if (/\[[A-Z]+_START\]/ && !beg) beg=NR; if (/\[[A-Z]+_END\]/) end=NR; next }
    FNR<beg || FNR>end
' file file
TEXT
TEXT
TEXT
TEXT

or keep the original condition but makes it's action "next" and add a default "print" for every other line to hit:

$ awk '
    NR==FNR { if (/\[[A-Z]+_START\]/ && !beg) beg=NR; if (/\[[A-Z]+_END\]/) end=NR; next }
    FNR>=beg && FNR<=end { next }
    { print }
' file file
TEXT
TEXT
TEXT
TEXT

Upvotes: 0

Jose Ricardo Bustos M.
Jose Ricardo Bustos M.

Reputation: 8164

A solution without awk, using grep

grep -Pzo '(?s)\[([A-Z]{3})_START\].*\n.*\[\1_END\]' file | sed 's/\x00/\n\n/'

you get,

[SUN_START]
[SUN_END]

[MON_START]
TEXT
[MON_END]

[TUE_START]
[TUE_END]

[WED_START]
TEXT
[WED_END]

*based in @albfan answer

Upvotes: 1

oliv
oliv

Reputation: 13249

You could use awk:

awk '/\[..._START\]/{p=1}/\[..._END\]/{print;p=0}p||!NF' file

The variable p is set when printing is needed. !NF allows to keep blank lines.

Upvotes: 0

Related Questions