How to parse username, ID or whole part using Ruby Regex in this sentence?
2
I have a sentences like this:
Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?
And what I want to is get @[Pratha](user:1) and @[John](user:3). Either their names and ids or just as texts as I quoted so that i can explode and parse name and id myself.
But there is an issue here. Names Pratha and John may include non-abc characters like ', ,, -, + , etc... But not and ()
What I tried so far:
c = ''
f = c.match(/(?:s|^)(?:@(?!(?:d+|w+?_|_w+?)(?:s([)|$)))(w+)(?=s|$)/i)
But no success.
ruby regex
asked Nov 23 '18 at 13:40
PrathaPratha
1907
add a comment |
2
I have a sentences like this:
Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?
And what I want to is get @[Pratha](user:1) and @[John](user:3). Either their names and ids or just as texts as I quoted so that i can explode and parse name and id myself.
But there is an issue here. Names Pratha and John may include non-abc characters like ', ,, -, + , etc... But not and ()
What I tried so far:
c = ''
f = c.match(/(?:s|^)(?:@(?!(?:d+|w+?_|_w+?)(?:s([)|$)))(w+)(?=s|$)/i)
But no success.
ruby regex
asked Nov 23 '18 at 13:40
PrathaPratha
1907
FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00
add a comment |
2
2
2
I have a sentences like this:
Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?
And what I want to is get @[Pratha](user:1) and @[John](user:3). Either their names and ids or just as texts as I quoted so that i can explode and parse name and id myself.
But there is an issue here. Names Pratha and John may include non-abc characters like ', ,, -, + , etc... But not and ()
What I tried so far:
c = ''
f = c.match(/(?:s|^)(?:@(?!(?:d+|w+?_|_w+?)(?:s([)|$)))(w+)(?=s|$)/i)
But no success.
ruby regex
asked Nov 23 '18 at 13:40
PrathaPratha
1907
I have a sentences like this:
Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?
And what I want to is get @[Pratha](user:1) and @[John](user:3). Either their names and ids or just as texts as I quoted so that i can explode and parse name and id myself.
But there is an issue here. Names Pratha and John may include non-abc characters like ', ,, -, + , etc... But not and ()
What I tried so far:
c = ''
f = c.match(/(?:s|^)(?:@(?!(?:d+|w+?_|_w+?)(?:s([)|$)))(w+)(?=s|$)/i)
But no success.
ruby regex
ruby regex
asked Nov 23 '18 at 13:40
PrathaPratha
1907
asked Nov 23 '18 at 13:40
PrathaPratha
1907
asked Nov 23 '18 at 13:40
PrathaPratha
1907
asked Nov 23 '18 at 13:40
PrathaPratha
1907
asked Nov 23 '18 at 13:40
PrathaPratha
1907
1907
FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00
add a comment |
FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00
FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00
FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00
add a comment |
3 Answers
3
active
oldest
votes
3
You may use
/@[([^]*)]([^()]*:(d+))/
See the regex demo
Details
@- a@char
[- a[
([^]*)- Group 1: 0+ chars other than[and]
]- a]char
(- a(char
[^()]*- 0+ chars other than(and)
:- a colon
(d+)- Group 2: 1 or more digits
)- a)char.
Sample Ruby code:
s = "Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?"
rx = /@[([^]*)]([^()]*:(d+))/
res = s.scan(rx)
puts res
# = > [["Pratha", "1"], ["John", "3"]]
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
I do not needuser:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?
– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably dropBbecause message (which user writes) maybe doesn't have a space before@For example:Hello@[Pratha](user:1)As you can see there is no space afterHello.
– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then(?:^|s)was a mistake in your pattern. I updated the code, regex and demos.
– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
add a comment |
1
"Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?".scan(/@.*?)/)
#⇒ ["@[Pratha](user:1)", "@[John](user:3)"]
Since the line is not coming from the user input, you might rely on that the part you are interested in starts with @ and ends with ).
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
add a comment |
1
You could use 2 capturing groups to get the names and the id's:
@[([^]]+)]([^:]+:([^)]+))
That will match
@Match literally
[Match [
([^]]+)1st capturing group which matches not ] 1+ times using a negated character class.
(Match literally
[^:]+:Match not :, then match :
([^)]+)2nd capturing group which matches not)1+ times
)Match)
Regex demo | Ruby demo
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we needBat start like Wiktor did? (to prevent start with a word)
– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained whatBdoes - it won't match@if it is inside of a word. Also, see my code snippet.
– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,Basserts the position wherebdoes not match.
– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use(?<!S)@[([^]*)](([^()]*))Demo to make sure what is on the left if not a non whitespace character.
– The fourth bird
Nov 23 '18 at 14:06
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You may use
/@[([^]*)]([^()]*:(d+))/
See the regex demo
Details
@- a@char
[- a[
([^]*)- Group 1: 0+ chars other than[and]
]- a]char
(- a(char
[^()]*- 0+ chars other than(and)
:- a colon
(d+)- Group 2: 1 or more digits
)- a)char.
Sample Ruby code:
s = "Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?"
rx = /@[([^]*)]([^()]*:(d+))/
res = s.scan(rx)
puts res
# = > [["Pratha", "1"], ["John", "3"]]
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
I do not needuser:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?
– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably dropBbecause message (which user writes) maybe doesn't have a space before@For example:Hello@[Pratha](user:1)As you can see there is no space afterHello.
– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then(?:^|s)was a mistake in your pattern. I updated the code, regex and demos.
– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
add a comment |
3
You may use
/@[([^]*)]([^()]*:(d+))/
See the regex demo
Details
@- a@char
[- a[
([^]*)- Group 1: 0+ chars other than[and]
]- a]char
(- a(char
[^()]*- 0+ chars other than(and)
:- a colon
(d+)- Group 2: 1 or more digits
)- a)char.
Sample Ruby code:
s = "Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?"
rx = /@[([^]*)]([^()]*:(d+))/
res = s.scan(rx)
puts res
# = > [["Pratha", "1"], ["John", "3"]]
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
I do not needuser:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?
– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably dropBbecause message (which user writes) maybe doesn't have a space before@For example:Hello@[Pratha](user:1)As you can see there is no space afterHello.
– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then(?:^|s)was a mistake in your pattern. I updated the code, regex and demos.
– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
add a comment |
3
3
3
You may use
/@[([^]*)]([^()]*:(d+))/
See the regex demo
Details
@- a@char
[- a[
([^]*)- Group 1: 0+ chars other than[and]
]- a]char
(- a(char
[^()]*- 0+ chars other than(and)
:- a colon
(d+)- Group 2: 1 or more digits
)- a)char.
Sample Ruby code:
s = "Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?"
rx = /@[([^]*)]([^()]*:(d+))/
res = s.scan(rx)
puts res
# = > [["Pratha", "1"], ["John", "3"]]
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
You may use
/@[([^]*)]([^()]*:(d+))/
See the regex demo
Details
@- a@char
[- a[
([^]*)- Group 1: 0+ chars other than[and]
]- a]char
(- a(char
[^()]*- 0+ chars other than(and)
:- a colon
(d+)- Group 2: 1 or more digits
)- a)char.
Sample Ruby code:
s = "Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?"
rx = /@[([^]*)]([^()]*:(d+))/
res = s.scan(rx)
puts res
# = > [["Pratha", "1"], ["John", "3"]]
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
edited Nov 23 '18 at 18:15
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
answered Nov 23 '18 at 13:48
Wiktor StribiżewWiktor Stribiżew
316k16134215
316k16134215
I do not needuser:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?
– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably dropBbecause message (which user writes) maybe doesn't have a space before@For example:Hello@[Pratha](user:1)As you can see there is no space afterHello.
– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then(?:^|s)was a mistake in your pattern. I updated the code, regex and demos.
– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
add a comment |
I do not needuser:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?
– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably dropBbecause message (which user writes) maybe doesn't have a space before@For example:Hello@[Pratha](user:1)As you can see there is no space afterHello.
– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then(?:^|s)was a mistake in your pattern. I updated the code, regex and demos.
– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
I do not need
user:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?– Pratha
Nov 23 '18 at 14:00
I do not need
user:. Just ID. Can we grab that only? And is there a specific difference between yours and @The fourth bird?– Pratha
Nov 23 '18 at 14:00
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
I think you misunderstand me. I just need name and ID. Like below. So output should be Pratha => 1, John => 3
– Pratha
Nov 23 '18 at 14:03
BTW, I will probably drop
B because message (which user writes) maybe doesn't have a space before @ For example: Hello@[Pratha](user:1) As you can see there is no space after Hello.– Pratha
Nov 23 '18 at 14:05
BTW, I will probably drop
B because message (which user writes) maybe doesn't have a space before @ For example: Hello@[Pratha](user:1) As you can see there is no space after Hello.– Pratha
Nov 23 '18 at 14:05
@Pratha Ok, then
(?:^|s) was a mistake in your pattern. I updated the code, regex and demos.– Wiktor Stribiżew
Nov 23 '18 at 14:08
@Pratha Ok, then
(?:^|s) was a mistake in your pattern. I updated the code, regex and demos.– Wiktor Stribiżew
Nov 23 '18 at 14:08
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
@CarySwoveland True, I updated the answer. I was playing with some Ruby code from another SO answer trying out various approaches and this block remained.
– Wiktor Stribiżew
Nov 23 '18 at 18:16
add a comment |
1
"Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?".scan(/@.*?)/)
#⇒ ["@[Pratha](user:1)", "@[John](user:3)"]
Since the line is not coming from the user input, you might rely on that the part you are interested in starts with @ and ends with ).
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
add a comment |
1
"Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?".scan(/@.*?)/)
#⇒ ["@[Pratha](user:1)", "@[John](user:3)"]
Since the line is not coming from the user input, you might rely on that the part you are interested in starts with @ and ends with ).
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
add a comment |
1
1
1
"Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?".scan(/@.*?)/)
#⇒ ["@[Pratha](user:1)", "@[John](user:3)"]
Since the line is not coming from the user input, you might rely on that the part you are interested in starts with @ and ends with ).
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
"Hello @[Pratha](user:1), did you see @[John](user:3)'s answer?".scan(/@.*?)/)
#⇒ ["@[Pratha](user:1)", "@[John](user:3)"]
Since the line is not coming from the user input, you might rely on that the part you are interested in starts with @ and ends with ).
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
answered Nov 23 '18 at 13:42
Aleksei MatiushkinAleksei Matiushkin
81.8k95591
81.8k95591
add a comment |
add a comment |
1
You could use 2 capturing groups to get the names and the id's:
@[([^]]+)]([^:]+:([^)]+))
That will match
@Match literally
[Match [
([^]]+)1st capturing group which matches not ] 1+ times using a negated character class.
(Match literally
[^:]+:Match not :, then match :
([^)]+)2nd capturing group which matches not)1+ times
)Match)
Regex demo | Ruby demo
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we needBat start like Wiktor did? (to prevent start with a word)
– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained whatBdoes - it won't match@if it is inside of a word. Also, see my code snippet.
– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,Basserts the position wherebdoes not match.
– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use(?<!S)@[([^]*)](([^()]*))Demo to make sure what is on the left if not a non whitespace character.
– The fourth bird
Nov 23 '18 at 14:06
add a comment |
1
You could use 2 capturing groups to get the names and the id's:
@[([^]]+)]([^:]+:([^)]+))
That will match
@Match literally
[Match [
([^]]+)1st capturing group which matches not ] 1+ times using a negated character class.
(Match literally
[^:]+:Match not :, then match :
([^)]+)2nd capturing group which matches not)1+ times
)Match)
Regex demo | Ruby demo
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we needBat start like Wiktor did? (to prevent start with a word)
– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained whatBdoes - it won't match@if it is inside of a word. Also, see my code snippet.
– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,Basserts the position wherebdoes not match.
– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use(?<!S)@[([^]*)](([^()]*))Demo to make sure what is on the left if not a non whitespace character.
– The fourth bird
Nov 23 '18 at 14:06
add a comment |
1
1
1
You could use 2 capturing groups to get the names and the id's:
@[([^]]+)]([^:]+:([^)]+))
That will match
@Match literally
[Match [
([^]]+)1st capturing group which matches not ] 1+ times using a negated character class.
(Match literally
[^:]+:Match not :, then match :
([^)]+)2nd capturing group which matches not)1+ times
)Match)
Regex demo | Ruby demo
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
You could use 2 capturing groups to get the names and the id's:
@[([^]]+)]([^:]+:([^)]+))
That will match
@Match literally
[Match [
([^]]+)1st capturing group which matches not ] 1+ times using a negated character class.
(Match literally
[^:]+:Match not :, then match :
([^)]+)2nd capturing group which matches not)1+ times
)Match)
Regex demo | Ruby demo
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
edited Nov 23 '18 at 14:24
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
answered Nov 23 '18 at 13:46
The fourth birdThe fourth bird
23k81427
23k81427
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we needBat start like Wiktor did? (to prevent start with a word)
– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained whatBdoes - it won't match@if it is inside of a word. Also, see my code snippet.
– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,Basserts the position wherebdoes not match.
– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use(?<!S)@[([^]*)](([^()]*))Demo to make sure what is on the left if not a non whitespace character.
– The fourth bird
Nov 23 '18 at 14:06
add a comment |
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we needBat start like Wiktor did? (to prevent start with a word)
– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained whatBdoes - it won't match@if it is inside of a word. Also, see my code snippet.
– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,Basserts the position wherebdoes not match.
– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use(?<!S)@[([^]*)](([^()]*))Demo to make sure what is on the left if not a non whitespace character.
– The fourth bird
Nov 23 '18 at 14:06
1
1
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
All answers are giving correct output, however, Wiktor's and yours parse user's separately. And your solution just parses ID which is perfect.
– Pratha
Nov 23 '18 at 13:54
Just a question, do we need
B at start like Wiktor did? (to prevent start with a word)– Pratha
Nov 23 '18 at 13:57
Just a question, do we need
B at start like Wiktor did? (to prevent start with a word)– Pratha
Nov 23 '18 at 13:57
@Pratha I have explained what
B does - it won't match @ if it is inside of a word. Also, see my code snippet.– Wiktor Stribiżew
Nov 23 '18 at 14:00
@Pratha I have explained what
B does - it won't match @ if it is inside of a word. Also, see my code snippet.– Wiktor Stribiżew
Nov 23 '18 at 14:00
It depends on what you would allow to match,
B asserts the position where b does not match.– The fourth bird
Nov 23 '18 at 14:01
It depends on what you would allow to match,
B asserts the position where b does not match.– The fourth bird
Nov 23 '18 at 14:01
@Pratha You could also use
(?<!S)@[([^]*)](([^()]*)) Demo to make sure what is on the left if not a non whitespace character.– The fourth bird
Nov 23 '18 at 14:06
@Pratha You could also use
(?<!S)@[([^]*)](([^()]*)) Demo to make sure what is on the left if not a non whitespace character.– The fourth bird
Nov 23 '18 at 14:06
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FWIW, that's an example of a library for your latest question, github.com/avivr/search-parser .
– estus
Dec 25 '18 at 19:00