Print text between two patterns

Question

I need to:

Print text between "hi" and "hello" patterns.

Patterns may be repeated and there may be multiple occurrences of hi..hello.

Source file:

hi aa bb cc
dd ee ff
hello xxxxxx
aaaa dddd cccc hi aaa bbb
ccc hello cccc fff

Expected output.

hi aa bb cc
dd ee ff
hello
hi aaa bbb
ccc hello

I tried with awk and sed -n commands whereas it displays all the line between the first occurrence of hi and the last occurrence of hello.

Answer 1

From the first hi to the first hello that follows.

grep

Using (GNU) grep and tr:

$ <infile grep -oPz "(?s)hi.*?hello" | tr '\0' '\n'
hi aa bb cc
dd ee ff
hello
hi aaa bbb
ccc hello

Descripttion:

• <infile Source file.
• grep -oPz Call grep to:
  • (-P) match a PCRE (Perl Compatible Regular Expression)
  • (-o) only print the matching part.
  • (-z) use a zero byte (a.k.a. NUL and a.k.a. \0) as line delimiter.
• "(?s) Make the PCRE dot (.) match also newlines.
• hi Starting with the string hi.
• .*? Match all characters that follow (non-greedy because of ?).
• hello" Up until the string hello is matched.
• | tr '\0' '\n' Convert the NULs (\0) bytes (from grep -z) to newlines.

sed

GNU sed:

<infile sed 's/hi/\n&/;s/[^\n]*\n//;s/\(hello\).*/\1/;/hi/,/hello/!d'

Or, for BSD sed, which doesn't allow \n on the right side of s///, you need to define a newline variable nl:

$ eval "$(printf "nl='\n'")"

And, then:

<infile sed 's/hi/\'"$nl"'&/;s/[^\n]*\n//;s/\(hello\).*/\1/;/hi/,/hello/!d'

Or; if you could write an explicit newline:

<infile sed 's/hi/\
&/;s/[^\n]*\n//;s/\(hello\).*/\1/;/hi/,/hello/!d'

Answer 2

grep -oPz "(?s)hi.*?hello" fileName

Explanation below.

cat tmp

Output ->

hi aa bb cc
dd ee ff
hello xxxxxx
aaaa dddd cccc hi aaa bbb
ccc hello cccc fff
00000000000
hi ff 
djd h
sdkf hello
dfj 

Solution:

grep -oPz "(?s)hi.*?hello" tmp

Output ->

hi aa bb cc
dd ee ff
hello
hi aaa bbb
ccc hello
hi ff 
djd h
sdkf hello

Params:

• -z this option tells grep to treat newlines as ordinary text characters, and look for null bytes to separate records.  In a text file with no null bytes, grep -z will treat the entire file as one line.

• (?s) activate PCRE_DOTALL, which means that '.' finds any character or newline.

• o Only print the matched part

• -P Interpret the pattern as a Perl-compatible regular expression (PCRE). This is highly experimental, particularly when combined with the -z (--null-data) option, and ‘grep -P’ may warn of unimplemented features.

• .*? For non-greedy match. Start searching for next match of hi when you find first instance of hello

  Source: https://www.gnu.org/software/grep/manual/grep.html

Answer 3

awk for giggles

#!/usr/bin/awk -f
{
    n=split($0, col, FS)
    for(i=1;i<=n;i++){
        (i==n)? sp="" : sp=FS
        if (col[i] ~ /hi/) p=1
        if (p == 1  && col[i] ~ /hello/) h=1
        if (p == 1) printf("%s%s",col[i],sp)
        if (h==1) p=0;h=0
    }
    printf "\n"
}



kapu@jake:$ cat SourceFile
hi aa bb cc
dd ee ff
hello xxxxxx
aaaa dddd cccc hi aaa bbb
ccc hello cccc fff
kapu@jake:$ script.awk SourceFile
hi aa bb cc
dd ee ff
hello
hi aaa bbb
ccc hello