Print lines after search of a pattern upto the end of the pattern

Question

From the many hits about searching a pattern and printing the lines until the end of the pattern, I've encountered an issue. The solution works if there is no other entry that is similar or has the common numbers in the pattern. Let me explain.

startpat=30.202.10.0  
endpat=31.202.10.0

inputfile is as follows:

30.202.10.0  
abcd  
efgh

31.202.10.0  
ijkl  
mnop

130.202.10.0  
qrst  
uvwx

awk -v startpat="${startpat}" -v endpat="${endpat}" '$0 ~ startpat{ x = 1} $0 ~ endpat{ x = 0}x' inputfile

I get the expected output. The problem arises when the data contains as an example 130.202.10.0. The result contains lines

abcd  
efgh  

qrst  
uvwx

How do I get as follows

abcd  
efgh

You should include lines like `31520251050` in your sample input/output as it's usually easy to match the text you want but much harder to not match similar text you don't want and that would let people test regexp vs string matching in addition to the full vs partial matching case you already have, `130.202.10.0`. See https://stackoverflow.com/q/65621325/1745001. — Ed Morton, Jun 30 '23 at 10:59
Do some lines in your input **really** have trailing blank characters at the end like in your example? — Ed Morton, Jun 30 '23 at 11:01

score 2 · Accepted Answer · answered Jun 30 '23 at 00:07

2

Current code:

$ awk -v startpat="${startpat}" -v endpat="${endpat}" '$0 ~ startpat{ x = 1} $0 ~ endpat{ x = 0}x' inputfile  
30.202.10.0
abcd
efgh

130.202.10.0
qrst
uvwx

Assuming the patterns (startpat and endpat) are not to be printed, and blank lines are not to be printed, one awk idea:

$ awk -v startpat="${startpat}" -v endpat="${endpat}" '$1==startpat {x=1;next} $1==endpat {x=0;next} x && NF>0' inputfile
abcd
efgh

answered Jun 30 '23 at 00:07

markp-fuso

28,790
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Thanks @markp_fuso, your solution worked!. – Aires69 Jun 30 '23 at 12:53

Grobu · Answer 2 · 2023-06-30T17:22:58.037

0

With SED :

startpat='30\.202\.10\.0'
endpat='31\.202\.10\.0'
sed -n -e "/^$startpat\$/!d; :loop; n; /^$endpat\$/d; p; b loop" INPUTFILE

Edited my answer to take into account Ed Morton's remark about dots matching any characters in the "startpat" / "endpat" variables. Thanks for pointing this out, I had totally overlooked that detail!

edited Jun 30 '23 at 17:22

answered Jun 30 '23 at 00:00

Grobu

599
1
11

score 0 · Answer 3 · edited Jul 03 '23 at 18:50

0

Using sed

$ sed -E "/\<${startpat//./\\.}\>/,/\<${endpat//./\\.}\>|^$/!d;//d" input_file
abcd
efgh

Edited my answer to take into account Ed Morton's remark about dots matching any characters in the "startpat" / "endpat" variables. Thanks for pointing this out, I had totally overlooked that detail!

edited Jul 03 '23 at 18:50

Grobu

599
1
11

answered Jun 30 '23 at 00:25

HatLess

10,622
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All those `.`s in startpat and endpat will match any characters. – Ed Morton Jun 30 '23 at 10:56

Ed Morton · Answer 4 · 2023-06-30T13:40:34.443

Using any awk handling 1 line at a time:

$ awk -v beg="$startpat" -v end="$endpat" '$1==end{f=0} f; $1==beg{f=1}' inputfile
abcd
efgh

or if you prefer to treat the input as empty-line-separated multi-line records:

$ awk -v beg="$startpat" -v end="$endpat" -v RS= -v ORS='\n\n' '$1==beg{f=1} $1==end{f=0} f{sub(/[^\n]+\n/,""); print}' inputfile
abcd
efgh

The above do full-string matching and so won't falsely match on substrings or regexp metachars. I'm testing for $1 instead of $0 to handle the trailing blanks in your input.

score 0 · Answer 5 · answered Jun 30 '23 at 18:46

This might work for you (GNU sed):

sed -nE '/^30\.202\.10\.0/{:a;n;/^31\.202\.10\.0/bb;/\S/H;ba;:b;g;s/.//p;z;x}' file

Match 30\.202\.10\.0 then collect further lines until another match of 31\.202\.10\.0 (dropping the first and last lines and only storing non-empty lines).

Remove an introduced starting newline and print the result.

Clean up ready for another match.

N.B. This solution remove empty lines, if all lines empty or not are required, use:

sed -nE '/^30\.202\.10\.0/{:a;n;/^31\.202\.10\.0/bb;H;ba;:b;g;s/.//p;z;x}' file

ufopilot · Answer 6 · 2023-07-03T16:03:10.120

0

startpat="30.202.10.0"  
endpat="31.202.10.0"

awk -v startpat="${startpat}" -v endpat="${endpat}" '
    BEGIN{gsub(/\./, "\\.", startpat); gsub(/\./, "\\.", endpat)}
    match($0, "^"startpat" *$"){
        found=1
        next
    }
    match($0, "^"endpat" *$"){
        exit
    }
    found && !/^$/
' file

abcd  
efgh

edited Jul 03 '23 at 16:03

answered Jun 30 '23 at 23:22

ufopilot

3,269
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1

@EdMorton: replaced with `\.` – ufopilot Jul 03 '23 at 16:04

score 0 · Answer 7 · answered Jul 01 '23 at 17:35

For pattern matching stuffs, perl is highly flexible & right choice. In Perl, using file slurp mode. Assuming the file content is available in aires.txt

perl -0777 -ne ' $start="30.202.10.0"; $end="31.202.10.0"; print $1 if m/$start(.+?)$end/s ' aires.txt

abcd  
efgh

Print lines after search of a pattern upto the end of the pattern

7 Answers7