Regex to search through a file for a set of parentheses with eight characters inside them, all 1s or 0s, with...
3
I'm trying to search through a file for a set of parentheses with eight characters inside them, all 1s or 0s, with at least one 1.
I currently am using the regex below, enumerating all possible such parentheses sets, i.e. brute forcing it.
Is there a better way to do this?
My regex:
(11111110)|(11111101)|(11111100)|(11111011)|(11111010)|(11111001)|(11111000)|...|(11111111) etc.
regex
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
add a comment |
3
I'm trying to search through a file for a set of parentheses with eight characters inside them, all 1s or 0s, with at least one 1.
I currently am using the regex below, enumerating all possible such parentheses sets, i.e. brute forcing it.
Is there a better way to do this?
My regex:
(11111110)|(11111101)|(11111100)|(11111011)|(11111010)|(11111001)|(11111000)|...|(11111111) etc.
regex
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
Yeah, usegrep -oP 'b(?=[01]{8}b)0*1[10]*'
– Wiktor Stribiżew
Nov 13 '18 at 19:10
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34
add a comment |
3
3
3
I'm trying to search through a file for a set of parentheses with eight characters inside them, all 1s or 0s, with at least one 1.
I currently am using the regex below, enumerating all possible such parentheses sets, i.e. brute forcing it.
Is there a better way to do this?
My regex:
(11111110)|(11111101)|(11111100)|(11111011)|(11111010)|(11111001)|(11111000)|...|(11111111) etc.
regex
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
I'm trying to search through a file for a set of parentheses with eight characters inside them, all 1s or 0s, with at least one 1.
I currently am using the regex below, enumerating all possible such parentheses sets, i.e. brute forcing it.
Is there a better way to do this?
My regex:
(11111110)|(11111101)|(11111100)|(11111011)|(11111010)|(11111001)|(11111000)|...|(11111111) etc.
regex
regex
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
edited Nov 13 '18 at 21:15
ttarchala
2,1071929
2,1071929
asked Nov 13 '18 at 19:04
anonanon
183
asked Nov 13 '18 at 19:04
anonanon
183
asked Nov 13 '18 at 19:04
anonanon
183
183
Yeah, usegrep -oP 'b(?=[01]{8}b)0*1[10]*'
– Wiktor Stribiżew
Nov 13 '18 at 19:10
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34
add a comment |
Yeah, usegrep -oP 'b(?=[01]{8}b)0*1[10]*'
– Wiktor Stribiżew
Nov 13 '18 at 19:10
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34
Yeah, use
grep -oP 'b(?=[01]{8}b)0*1[10]*'– Wiktor Stribiżew
Nov 13 '18 at 19:10
Yeah, use
grep -oP 'b(?=[01]{8}b)0*1[10]*'– Wiktor Stribiżew
Nov 13 '18 at 19:10
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34
add a comment |
3 Answers
3
active
oldest
votes
3
Use
grep -oP '((?=[01]{8}))0*1[10]*)' file
See the regex demo.
Details
-o- outputs matches rather than matching lines
P- enables PCRE regex engine
Pattern
(- a(char
(?=[01]{8}))- a positive lookahead that requires eight0or1chars up to to a)
0*- 0+ zeros
1- a1char
[10]*- 0 or more zeros or ones
)- a).
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
add a comment |
1
You may use this regex,
((?=[10]{8})(?=.*1.*).{8})
Explanation:
(--> Matches a literal((starting brace)
(?=[10]{8})--> Look ahead ensuring next eight characters are composed of zero and one only
(?=.*1.*)--> Look ahead ensuring the presence of at least one '1' character
.{8}--> matches exactly eight characters
)--> Matches a literal)(closing brace)
Demo
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
add a comment |
0
This regex could also work:
(?!(00000000))(d{8})
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
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
Use
grep -oP '((?=[01]{8}))0*1[10]*)' file
See the regex demo.
Details
-o- outputs matches rather than matching lines
P- enables PCRE regex engine
Pattern
(- a(char
(?=[01]{8}))- a positive lookahead that requires eight0or1chars up to to a)
0*- 0+ zeros
1- a1char
[10]*- 0 or more zeros or ones
)- a).
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
add a comment |
3
Use
grep -oP '((?=[01]{8}))0*1[10]*)' file
See the regex demo.
Details
-o- outputs matches rather than matching lines
P- enables PCRE regex engine
Pattern
(- a(char
(?=[01]{8}))- a positive lookahead that requires eight0or1chars up to to a)
0*- 0+ zeros
1- a1char
[10]*- 0 or more zeros or ones
)- a).
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
add a comment |
3
3
3
Use
grep -oP '((?=[01]{8}))0*1[10]*)' file
See the regex demo.
Details
-o- outputs matches rather than matching lines
P- enables PCRE regex engine
Pattern
(- a(char
(?=[01]{8}))- a positive lookahead that requires eight0or1chars up to to a)
0*- 0+ zeros
1- a1char
[10]*- 0 or more zeros or ones
)- a).
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
Use
grep -oP '((?=[01]{8}))0*1[10]*)' file
See the regex demo.
Details
-o- outputs matches rather than matching lines
P- enables PCRE regex engine
Pattern
(- a(char
(?=[01]{8}))- a positive lookahead that requires eight0or1chars up to to a)
0*- 0+ zeros
1- a1char
[10]*- 0 or more zeros or ones
)- a).
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
edited Nov 13 '18 at 19:36
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
answered Nov 13 '18 at 19:30
Wiktor StribiżewWiktor Stribiżew
314k16133212
314k16133212
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
add a comment |
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
1
1
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
This works great too. Of course, as it is by Wiktor "The Great Regexer" :)
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:37
add a comment |
1
You may use this regex,
((?=[10]{8})(?=.*1.*).{8})
Explanation:
(--> Matches a literal((starting brace)
(?=[10]{8})--> Look ahead ensuring next eight characters are composed of zero and one only
(?=.*1.*)--> Look ahead ensuring the presence of at least one '1' character
.{8}--> matches exactly eight characters
)--> Matches a literal)(closing brace)
Demo
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
add a comment |
1
You may use this regex,
((?=[10]{8})(?=.*1.*).{8})
Explanation:
(--> Matches a literal((starting brace)
(?=[10]{8})--> Look ahead ensuring next eight characters are composed of zero and one only
(?=.*1.*)--> Look ahead ensuring the presence of at least one '1' character
.{8}--> matches exactly eight characters
)--> Matches a literal)(closing brace)
Demo
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
add a comment |
1
1
1
You may use this regex,
((?=[10]{8})(?=.*1.*).{8})
Explanation:
(--> Matches a literal((starting brace)
(?=[10]{8})--> Look ahead ensuring next eight characters are composed of zero and one only
(?=.*1.*)--> Look ahead ensuring the presence of at least one '1' character
.{8}--> matches exactly eight characters
)--> Matches a literal)(closing brace)
Demo
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
You may use this regex,
((?=[10]{8})(?=.*1.*).{8})
Explanation:
(--> Matches a literal((starting brace)
(?=[10]{8})--> Look ahead ensuring next eight characters are composed of zero and one only
(?=.*1.*)--> Look ahead ensuring the presence of at least one '1' character
.{8}--> matches exactly eight characters
)--> Matches a literal)(closing brace)
Demo
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
answered Nov 13 '18 at 19:28
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
7,5282927
7,5282927
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
add a comment |
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
Thanks for your help, however that still finds the string (00000000) which I was hoping to avoid, as its the base case
– anon
Nov 13 '18 at 19:31
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
It doesn't match (00000000). Check this regex101.com/r/lr8haq/2
– Pushpesh Kumar Rajwanshi
Nov 13 '18 at 19:32
add a comment |
0
This regex could also work:
(?!(00000000))(d{8})
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
add a comment |
0
This regex could also work:
(?!(00000000))(d{8})
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
add a comment |
0
0
0
This regex could also work:
(?!(00000000))(d{8})
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
This regex could also work:
(?!(00000000))(d{8})
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
edited Nov 13 '18 at 20:18
Tiago Martins Peres
2,21341935
2,21341935
answered Nov 13 '18 at 20:11
ecaglecagl
1
answered Nov 13 '18 at 20:11
ecaglecagl
1
answered Nov 13 '18 at 20:11
ecaglecagl
1
1
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
add a comment |
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
See Lookarounds (Usually) Want to be Anchored.
– Wiktor Stribiżew
Nov 13 '18 at 21:24
add a comment |
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Yeah, use
grep -oP 'b(?=[01]{8}b)0*1[10]*'– Wiktor Stribiżew
Nov 13 '18 at 19:10
See my answer below, I modified my comment as I did not realize you wanted to match these binary data inside parentheses.
– Wiktor Stribiżew
Nov 13 '18 at 19:34