Using Sed to Print Between Two Patterns

How to print lines between two patterns, inclusive or exclusive (in sed, AWK or Perl)?

Print lines between PAT1 and PAT2

$ awk '/PAT1/,/PAT2/' file
PAT1
3    - first block
4
PAT2
PAT1
7    - second block
PAT2
PAT1
10    - third block

Or, using variables:

awk '/PAT1/{flag=1} flag; /PAT2/{flag=0}' file

How does this work?

/PAT1/ matches lines having this text, as well as /PAT2/ does.
/PAT1/{flag=1} sets the flag when the text PAT1 is found in a line.
/PAT2/{flag=0} unsets the flag when the text PAT2 is found in a line.
flag is a pattern with the default action, which is to print $0: if flag is equal 1 the line is printed. This way, it will print all those lines occurring from the time PAT1 occurs and up to the next PAT2 is seen. This will also print the lines from the last match of PAT1 up to the end of the file.

Print lines between PAT1 and PAT2 - not including PAT1 and PAT2

$ awk '/PAT1/{flag=1; next} /PAT2/{flag=0} flag' file
3    - first block
4
7    - second block
10    - third block

This uses next to skip the line that contains PAT1 in order to avoid this being printed.

This call to next can be dropped by reshuffling the blocks: awk '/PAT2/{flag=0} flag; /PAT1/{flag=1}' file.

Print lines between PAT1 and PAT2 - including PAT1

$ awk '/PAT1/{flag=1} /PAT2/{flag=0} flag' file
PAT1
3    - first block
4
PAT1
7    - second block
PAT1
10    - third block

By placing flag at the very end, it triggers the action that was set on either PAT1 or PAT2: to print on PAT1, not to print on PAT2.

Print lines between PAT1 and PAT2 - including PAT2

$ awk 'flag; /PAT1/{flag=1} /PAT2/{flag=0}' file
3    - first block
4
PAT2
7    - second block
PAT2
10    - third block

By placing flag at the very beginning, it triggers the action that was set previously and hence print the closing pattern but not the starting one.

Print lines between PAT1 and PAT2 - excluding lines from the last PAT1 to the end of file if no other PAT2 occurs

This is based on a solution by Ed Morton.

awk 'flag{
        if (/PAT2/)
           {printf "%s", buf; flag=0; buf=""}
        else
            buf = buf $0 ORS
     }
     /PAT1/ {flag=1}' file

As a one-liner:

$ awk 'flag{ if (/PAT2/){printf "%s", buf; flag=0; buf=""} else buf = buf $0 ORS}; /PAT1/{flag=1}' file
3    - first block
4
7    - second block

# note the lack of third block, since no other PAT2 happens after it

This keeps all the selected lines in a buffer that gets populated from the moment PAT1 is found. Then, it keeps being filled with the following lines until PAT2 is found. In that point, it prints the stored content and empties the buffer.

Print all lines between two patterns, exclusive, first instance only (in sed, AWK or Perl)

With awk (assumes that PATTERN1 and PATTERN2 are always present in pairs and either of them do not occur inside a pair)

$ cat ip.txt
aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
fff
PATTERN1
ggg
hhh
iii
PATTERN2
jjj

$ awk '/PATTERN2/{exit} f; /PATTERN1/{f=1}' ip.txt
bbb
ccc
ddd

/PATTERN1/{f=1} set flag if /PATTERN1/ is matched
/PATTERN2/{exit} exit if /PATTERN2/ is matched
f; print input line if flag is set

Generic solution, where the block required can be specified

$ awk -v b=1 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
bbb
ccc
ddd
$ awk -v b=2 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
2
46

how to print Lines Between Two Patterns in file using SED or AWK?

You may use this sed:

sed -n '/MULTIPLE-RESOURCES/,/^###$/ { /###$/!p; }' file

### MULTIPLE-RESOURCES

#### Viewing Resource Information

> kubectl get svc, po
> kubectl get deploy, no
> kubectl get all
> kubectl get all --all-namespaces

## KUBECTL

How can I use sed to match between two patterns and get the next line after the second pattern?

You can use N (syntax here is based on GNU sed)

$ sed -n '/PATTERN1/,/PATTERN2/{/PATTERN2/N; p}' ip.txt 
PATTERN1
B
PATTERN2
C

Using awk

$ awk '/PATTERN1/{f=1} f || (c && c--); /PATTERN2/{f=0; c=1}' ip.txt
PATTERN1
B
PATTERN2
C

which you can generalize using:

awk -v n=2 '/PATTERN1/{f=1} f || (c && c--); /PATTERN2/{f=0; c=n}'

Further Reading:

How to print lines between two patterns, inclusive or exclusive
Printing with sed or awk a line following a matching pattern

Output text between two patterns on the same line using sed command

For your current input you may use this sed:

sed 's/.*begin \(.*\) end.*/\1/' file

not sure what is wrong

Difference is use of .* after end that matches text after last end and discards in substitution.

However for your 2nd part if there are two end words, sed command won't work correctly as it will find last end due to greedy matching of .*.

e.g if your input is:

this is begin not sure what is wrong end and why not end

Then following awk would work better:

awk -F 'begin | end' '{print $2}' file

not sure what is wrong

Quit after printing between two patterns using sed

What you need is:

sed -n '/pattern1/,/pattern2/{/pattern2/{q};p}' file

Which has the form:

sed -n '/PAT1/,/PAT2/{/PAT2/{q};p}'

Which breaks down as

sed -n '/PAT1/,/PAT2/ - locate the range between PAT1 and PAT2 and suppress printing;
/PAT2/{q}; - if it matches PAT2 quit (done);
p - print all lines that fell within /PAT1/,/PAT2/ before quit was reached.

Example Use/Output

With your input in file, you would get:

$ sed -n '/pattern1/,/pattern2/{/pattern2/{q};p}' file
pattern1
example1

SED or AWK command that print the value between two patterns

awk -F'[=&]' '{print $2}' file 
1234567890

sed command to print lines between two patterns

You can do it with awk: awk '/patt1/{flag=1}/patt2/{flag=0}flag' input_file

If input_file is:

awk '/222/{flag=1}/444/{flag=0}flag' input_file

gives:

222
333

Using sed to print lines between 2 patterns

I came up to this command

sed -n '1,/^Index:/{/^Index:/!d;}; /^Index:/{x;/^$/!p;n;n;}; H; ${g;p;};'

It deletes from start till the first Index: line
Then it saves into a holding buffer everything from Index: till next Index: excluding ======= line, which is assumed to follow Index: immediately
When Index: line is met, it prints the content of the holding buffer if it's not empty

When the end of the file reached it prints the content of the holding buffer

$ cat /tmp/test
First line
Index: <filepath>
===================================================================
<lines to print>
<lines to print>
<lines to print>
Index: <filepath>
===================================================================
<lines to print>
<lines to print>
<lines to print>

$ sed -n '1,/^Index:/{/^Index:/!d;}; /^Index:/{x;/^$/!p;n;n;}; H; ${g;p;};' \
/tmp/test
Index: <filepath>
<lines to print>
<lines to print>
<lines to print>
Index: <filepath>
<lines to print>
<lines to print>
<lines to print>

But as David mentioned, it can be shorten and then it's just a simple

sed '1,/^Index:/{/^Index:/!d;}; /^=/d;' /tmp/test

which is just the same for lines before the first Index and then just removing lines starting with =