grep sentence between (and including) two patterns

up vote
2
down vote

favorite

I want to extract sentences which start with

https://www.instagram.com/p/

and end with

For example, I want to extract the following without the x's

ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—

I have already tried

grep "https://www.instagram.com/p/*/"

However, it is not working.

edited Aug 6 at 16:38

SivaPrasath

3,68311636

asked Aug 6 at 16:08

Yusuke Otsubo

132

add a commentÂ |Â

up vote
2
down vote

favorite

I want to extract sentences which start with

https://www.instagram.com/p/

and end with

For example, I want to extract the following without the x's

ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—

I have already tried

grep "https://www.instagram.com/p/*/"

However, it is not working.

edited Aug 6 at 16:38

SivaPrasath

3,68311636

asked Aug 6 at 16:08

Yusuke Otsubo

132

add a commentÂ |Â

up vote
2
down vote

favorite

I want to extract sentences which start with

https://www.instagram.com/p/

and end with

For example, I want to extract the following without the x's

ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—

I have already tried

grep "https://www.instagram.com/p/*/"

However, it is not working.

edited Aug 6 at 16:38

SivaPrasath

3,68311636

asked Aug 6 at 16:08

Yusuke Otsubo

132

I want to extract sentences which start with

https://www.instagram.com/p/

and end with

For example, I want to extract the following without the x's

ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—

I have already tried

grep "https://www.instagram.com/p/*/"

However, it is not working.

edited Aug 6 at 16:38

SivaPrasath

3,68311636

asked Aug 6 at 16:08

Yusuke Otsubo

132

edited Aug 6 at 16:38

SivaPrasath

3,68311636

edited Aug 6 at 16:38

SivaPrasath

3,68311636

edited Aug 6 at 16:38

SivaPrasath

3,68311636

asked Aug 6 at 16:08

Yusuke Otsubo

132

asked Aug 6 at 16:08

Yusuke Otsubo

132

asked Aug 6 at 16:08

Yusuke Otsubo

132

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
1
down vote

accepted

Try the following regular expression, https://www.instagram.com/p/[^/]+/

#!/bin/bash
data="ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—"
echo "$data" | grep -o 'https://www.instagram.com/p/[^/]+/'

The magic part is [^/]+/, it grabs everything up to the next forward slash.

Sample output from the above script.

zb@server ~ $ ./tmp.sh 
https://www.instagram.com/p/BRhNDg5jne7/

answered Aug 6 at 16:18

Zachary Brady

3,376831

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

add a commentÂ |Â

up vote
3
down vote

Using grep :

echo "ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—" | grep -Po "(?s)(http(.*?)(/p/.*/|/Z))"

output:

https://www.instagram.com/p/BRhNDg5jne7/

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

add a commentÂ |Â

up vote
0
down vote

no need of perl regex You can try :

grep -o "https://www.instagram.com/.*/"

answered Aug 6 at 18:08

ctac_

986116

1

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

add a commentÂ |Â

up vote
0
down vote

EDIT: since the question had some changes since I posted my answer, so did my understanding of it.

If all your rows have the pattern xxxx, then all you gotta do is a regex replace with sed. I.e.:

sed 's/xxxx*//g'

If you first need to grep the rows, then pipe sed after grep. I.e.:

grep "https://www.instagram.com/p/" | sed 's/xxxx*//g'

Depending of the real pattern you have, this approach may or may not be of use.

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
1
down vote

accepted

Try the following regular expression, https://www.instagram.com/p/[^/]+/

#!/bin/bash
data="ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—"
echo "$data" | grep -o 'https://www.instagram.com/p/[^/]+/'

The magic part is [^/]+/, it grabs everything up to the next forward slash.

Sample output from the above script.

zb@server ~ $ ./tmp.sh 
https://www.instagram.com/p/BRhNDg5jne7/

answered Aug 6 at 16:18

Zachary Brady

3,376831

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

add a commentÂ |Â

up vote
1
down vote

accepted

Try the following regular expression, https://www.instagram.com/p/[^/]+/

#!/bin/bash
data="ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—"
echo "$data" | grep -o 'https://www.instagram.com/p/[^/]+/'

The magic part is [^/]+/, it grabs everything up to the next forward slash.

Sample output from the above script.

zb@server ~ $ ./tmp.sh 
https://www.instagram.com/p/BRhNDg5jne7/

answered Aug 6 at 16:18

Zachary Brady

3,376831

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

add a commentÂ |Â

up vote
1
down vote

accepted

Try the following regular expression, https://www.instagram.com/p/[^/]+/

#!/bin/bash
data="ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—"
echo "$data" | grep -o 'https://www.instagram.com/p/[^/]+/'

The magic part is [^/]+/, it grabs everything up to the next forward slash.

Sample output from the above script.

zb@server ~ $ ./tmp.sh 
https://www.instagram.com/p/BRhNDg5jne7/

answered Aug 6 at 16:18

Zachary Brady

3,376831

Try the following regular expression, https://www.instagram.com/p/[^/]+/

#!/bin/bash
data="ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—"
echo "$data" | grep -o 'https://www.instagram.com/p/[^/]+/'

The magic part is [^/]+/, it grabs everything up to the next forward slash.

Sample output from the above script.

zb@server ~ $ ./tmp.sh 
https://www.instagram.com/p/BRhNDg5jne7/

answered Aug 6 at 16:18

Zachary Brady

3,376831

answered Aug 6 at 16:18

Zachary Brady

3,376831

answered Aug 6 at 16:18

Zachary Brady

3,376831

answered Aug 6 at 16:18

Zachary Brady

3,376831

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

add a commentÂ |Â

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

It works! Thank you!
â€“Â Yusuke Otsubo
yesterday

add a commentÂ |Â

up vote
3
down vote

Using grep :

echo "ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—" | grep -Po "(?s)(http(.*?)(/p/.*/|/Z))"

output:

https://www.instagram.com/p/BRhNDg5jne7/

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

add a commentÂ |Â

up vote
3
down vote

Using grep :

echo "ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—" | grep -Po "(?s)(http(.*?)(/p/.*/|/Z))"

output:

https://www.instagram.com/p/BRhNDg5jne7/

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

add a commentÂ |Â

up vote
3
down vote

Using grep :

echo "ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—" | grep -Po "(?s)(http(.*?)(/p/.*/|/Z))"

output:

https://www.instagram.com/p/BRhNDg5jne7/

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

Using grep :

echo "ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—https://www.instagram.com/p/BRhNDg5jne7/ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—ÃƒÂ—" | grep -Po "(?s)(http(.*?)(/p/.*/|/Z))"

output:

https://www.instagram.com/p/BRhNDg5jne7/

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

edited Aug 6 at 18:05

answered Aug 6 at 16:37

SivaPrasath

3,68311636

answered Aug 6 at 16:37

SivaPrasath

3,68311636

answered Aug 6 at 16:37

SivaPrasath

3,68311636

add a commentÂ |Â

up vote
0
down vote

no need of perl regex You can try :

grep -o "https://www.instagram.com/.*/"

answered Aug 6 at 18:08

ctac_

986116

1

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

add a commentÂ |Â

up vote
0
down vote

no need of perl regex You can try :

grep -o "https://www.instagram.com/.*/"

answered Aug 6 at 18:08

ctac_

986116

1

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

add a commentÂ |Â

up vote
0
down vote

no need of perl regex You can try :

grep -o "https://www.instagram.com/.*/"

answered Aug 6 at 18:08

ctac_

986116

no need of perl regex You can try :

grep -o "https://www.instagram.com/.*/"

answered Aug 6 at 18:08

ctac_

986116

answered Aug 6 at 18:08

ctac_

986116

answered Aug 6 at 18:08

ctac_

986116

answered Aug 6 at 18:08

ctac_

986116

1

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

add a commentÂ |Â

1

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

You probably want a non-greedy match, or something like https://www.instagram.com/foo/ bar baz other stuff/ will match the entire thing. You can do it by passing -P and changing .* to .*?
â€“Â Michael Mrozekâ™¦
Aug 6 at 19:04

add a commentÂ |Â

up vote
0
down vote

EDIT: since the question had some changes since I posted my answer, so did my understanding of it.

If all your rows have the pattern xxxx, then all you gotta do is a regex replace with sed. I.e.:

sed 's/xxxx*//g'

If you first need to grep the rows, then pipe sed after grep. I.e.:

grep "https://www.instagram.com/p/" | sed 's/xxxx*//g'

Depending of the real pattern you have, this approach may or may not be of use.

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

add a commentÂ |Â

up vote
0
down vote

EDIT: since the question had some changes since I posted my answer, so did my understanding of it.

If all your rows have the pattern xxxx, then all you gotta do is a regex replace with sed. I.e.:

sed 's/xxxx*//g'

If you first need to grep the rows, then pipe sed after grep. I.e.:

grep "https://www.instagram.com/p/" | sed 's/xxxx*//g'

Depending of the real pattern you have, this approach may or may not be of use.

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

add a commentÂ |Â

up vote
0
down vote

EDIT: since the question had some changes since I posted my answer, so did my understanding of it.

If all your rows have the pattern xxxx, then all you gotta do is a regex replace with sed. I.e.:

sed 's/xxxx*//g'

If you first need to grep the rows, then pipe sed after grep. I.e.:

grep "https://www.instagram.com/p/" | sed 's/xxxx*//g'

Depending of the real pattern you have, this approach may or may not be of use.

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

EDIT: since the question had some changes since I posted my answer, so did my understanding of it.

If all your rows have the pattern xxxx, then all you gotta do is a regex replace with sed. I.e.:

sed 's/xxxx*//g'

If you first need to grep the rows, then pipe sed after grep. I.e.:

grep "https://www.instagram.com/p/" | sed 's/xxxx*//g'

Depending of the real pattern you have, this approach may or may not be of use.

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

edited Aug 6 at 20:48

answered Aug 6 at 16:25

Juanse Albuja

answered Aug 6 at 16:25

Juanse Albuja

answered Aug 6 at 16:25

Juanse Albuja

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

add a commentÂ |Â

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

hope you are removing, instead of printing.
â€“Â SivaPrasath
Aug 6 at 17:27

@SivaPrasath please see my edited answer.
â€“Â Juanse Albuja
Aug 6 at 20:48

add a commentÂ |Â

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