Wildcards in RegEx Haystack (target)?
I want to search for a shorter string in a longer string. Can the longer string (haystack/target) be setup with wildcards?
e.g.) I want to look for 1234 in a string 1234578. Sometimes the long string will be 1@345678 or 12@45678 due to I/O errors. How do I setup the RegEx to match 1234 to 12345678 and any location of the @ symbol anywhere in the long string?
Reason: I have a scanner and it sometimes does not read one character correctly. It replaces that character with a @ symbol. How do I setup my RegEx so that I can have it ignore that @ character and match it?
regex
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
add a comment |
I want to search for a shorter string in a longer string. Can the longer string (haystack/target) be setup with wildcards?
e.g.) I want to look for 1234 in a string 1234578. Sometimes the long string will be 1@345678 or 12@45678 due to I/O errors. How do I setup the RegEx to match 1234 to 12345678 and any location of the @ symbol anywhere in the long string?
Reason: I have a scanner and it sometimes does not read one character correctly. It replaces that character with a @ symbol. How do I setup my RegEx so that I can have it ignore that @ character and match it?
regex
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06
add a comment |
I want to search for a shorter string in a longer string. Can the longer string (haystack/target) be setup with wildcards?
e.g.) I want to look for 1234 in a string 1234578. Sometimes the long string will be 1@345678 or 12@45678 due to I/O errors. How do I setup the RegEx to match 1234 to 12345678 and any location of the @ symbol anywhere in the long string?
Reason: I have a scanner and it sometimes does not read one character correctly. It replaces that character with a @ symbol. How do I setup my RegEx so that I can have it ignore that @ character and match it?
regex
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
I want to search for a shorter string in a longer string. Can the longer string (haystack/target) be setup with wildcards?
e.g.) I want to look for 1234 in a string 1234578. Sometimes the long string will be 1@345678 or 12@45678 due to I/O errors. How do I setup the RegEx to match 1234 to 12345678 and any location of the @ symbol anywhere in the long string?
Reason: I have a scanner and it sometimes does not read one character correctly. It replaces that character with a @ symbol. How do I setup my RegEx so that I can have it ignore that @ character and match it?
regex
regex
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
asked Nov 12 '18 at 9:57
NelsonNelson
1,159817
1,159817
@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06
add a comment |
@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06
@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06
add a comment |
1 Answer
1
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votes
Since I have programmatic control of the parsing process, I can programmatically modify the search string.
So instead of /1234/, it'll end up being /(1|@)(2|@)(3|@)(4|@)/
It's... really ugly, but that's my current solution.
The process isn't particularly difficult. I'll use regex on the search string by using /(.)/gm and replace it with ($1|@) so source 1234 becomes (1|@)(2|@)(3|@)(4|@) and search with that.
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like(?:1234|@234|1@34|12@4|123@)Demo
– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn1234into(?:1234|@234|1@34|12@4|123@)?
– Nelson
Nov 13 '18 at 1:28
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Since I have programmatic control of the parsing process, I can programmatically modify the search string.
So instead of /1234/, it'll end up being /(1|@)(2|@)(3|@)(4|@)/
It's... really ugly, but that's my current solution.
The process isn't particularly difficult. I'll use regex on the search string by using /(.)/gm and replace it with ($1|@) so source 1234 becomes (1|@)(2|@)(3|@)(4|@) and search with that.
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like(?:1234|@234|1@34|12@4|123@)Demo
– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn1234into(?:1234|@234|1@34|12@4|123@)?
– Nelson
Nov 13 '18 at 1:28
add a comment |
Since I have programmatic control of the parsing process, I can programmatically modify the search string.
So instead of /1234/, it'll end up being /(1|@)(2|@)(3|@)(4|@)/
It's... really ugly, but that's my current solution.
The process isn't particularly difficult. I'll use regex on the search string by using /(.)/gm and replace it with ($1|@) so source 1234 becomes (1|@)(2|@)(3|@)(4|@) and search with that.
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like(?:1234|@234|1@34|12@4|123@)Demo
– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn1234into(?:1234|@234|1@34|12@4|123@)?
– Nelson
Nov 13 '18 at 1:28
add a comment |
Since I have programmatic control of the parsing process, I can programmatically modify the search string.
So instead of /1234/, it'll end up being /(1|@)(2|@)(3|@)(4|@)/
It's... really ugly, but that's my current solution.
The process isn't particularly difficult. I'll use regex on the search string by using /(.)/gm and replace it with ($1|@) so source 1234 becomes (1|@)(2|@)(3|@)(4|@) and search with that.
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
Since I have programmatic control of the parsing process, I can programmatically modify the search string.
So instead of /1234/, it'll end up being /(1|@)(2|@)(3|@)(4|@)/
It's... really ugly, but that's my current solution.
The process isn't particularly difficult. I'll use regex on the search string by using /(.)/gm and replace it with ($1|@) so source 1234 becomes (1|@)(2|@)(3|@)(4|@) and search with that.
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
answered Nov 12 '18 at 9:59
NelsonNelson
1,159817
1,159817
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like(?:1234|@234|1@34|12@4|123@)Demo
– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn1234into(?:1234|@234|1@34|12@4|123@)?
– Nelson
Nov 13 '18 at 1:28
add a comment |
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like(?:1234|@234|1@34|12@4|123@)Demo
– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn1234into(?:1234|@234|1@34|12@4|123@)?
– Nelson
Nov 13 '18 at 1:28
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like
(?:1234|@234|1@34|12@4|123@) Demo– The fourth bird
Nov 12 '18 at 17:31
That way they can all be @ symbols. If you only want to match 1 @ out of 1-4, you could use an alternation like
(?:1234|@234|1@34|12@4|123@) Demo– The fourth bird
Nov 12 '18 at 17:31
@Thefourthbird Hmm, you got any idea on how to turn
1234 into (?:1234|@234|1@34|12@4|123@)?– Nelson
Nov 13 '18 at 1:28
@Thefourthbird Hmm, you got any idea on how to turn
1234 into (?:1234|@234|1@34|12@4|123@)?– Nelson
Nov 13 '18 at 1:28
add a comment |
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@Toto If I had 1@345678, and I remove it, it becomes 1345678 and will never match 1234. I'm not sure how I can make it work that way.
– Nelson
Nov 12 '18 at 13:57
You're right, I misread ;(
– Toto
Nov 12 '18 at 14:06