Match pattern anywhere in string? - regex

I want to match the following pattern:
Exxxx49 (where x is a digit 0-9)
For example, E123449abcdefgh, abcdefE123449987654321 are both valid. I.e., I need to match the pattern anywhere in a string.
I am using:
^*E[0-9]{4}49*$
But it only matches E123449.
How can I allow any amount of characters in front or after the pattern?

Remove the ^ and $ to search anywhere in the string.
In your case the * are probably not what you intended; E[0-9]{4}49 should suffice. This will find an E, followed by four digits, followed by a 4 and a 9, anywhere in the string.

I would go for
^.*E[0-9]{4}49.*$
EDIT:
since it fullfills all requirements state by OP.
"[match] Exxxx49 (where x is digit 0-9)"
"allow for any amount of characters in front or after pattern"
It will match
^.* everything from, including the beginning of the line
E[0-9]{4}49 the requested pattern
.*$ everthing after the pattern, including the the end of the line

Your original regex had a regex pattern syntax error at the first *. Fix it and change it to this:
.*E\d{4}49.*
This pattern is for matching in engines (most engines) that are anchored, like Java. Since you forgot to specify a language.
.* matches any number of sequences. As it surrounds the match, this will match the entire string as long as this match is located in the string.
Here is a regex demo!

Just simply use this:
E[0-9]{4}49

How do I allow for any amount of characters in front or after pattern? but it only matches E123449
Use global flag /E\d{4}49/g if supported by the language
OR
Try with capturing groups (E\d{4}49)+ that is grouped by enclosing inside parenthesis (...)
Here is online demo

Related

Why the character ^ is required in an regex ^(?!.*?spam) to filter strings?

I try to filter strings, that don't contain word "spam".
I use the regex from here!
But I can't understand why I need the symbol ^ at the start of expression. I know that it signs the start of regex but I do not understand why it doesn't work without ^ in my case?
UPD. All the answers hereunder are very usefull.
It's completely clear now. Thank you!
The regex (?!.*?spam) matches a position in a string that is not followed by something matching .*?spam.
Every single string has such a position, because if nothing else, the very end of the string is certainly not followed by anything matching .*?spam.
So every single string contains a match for the regex (?!.*?spam).
The anchor ^ in ^(?!.*?spam) restricts the regex, so that it only matches strings where the very beginning of the string isn't followed by anything matching .*?spam — i.e., strings that don't contain spam at all (or anywhere in the first line, at least, depending on whether . matches newlines).
The lookahead is a zero-width assertion (that is, it ensures a position in your string). In your case it is a negative lookahead making sure that not "zero more characters, followed by the word spam" are following. This is true for a couple of positions in your string, see a demo on regex101.com without the anchor.
With the anchor the matching process starts right at the very beginning, so the whole string is analyzed, see the altered demo on regex101.com as well.

How to find a particular string

Im using Visual Studio 2017 and in a long long text file Im searching for a particular function but unable to find
here's what the regex Im using
c\.CreateMap\<(\w)+\,\s+Address\>
and I want to in these
c.CreateMap<ClientAddress, Address>()
c.CreateMap<Responses.SiteAddress, Data.Address>()
and so on.
As soon as I add "Address" in the regex it stops matching any.
what am I doing wrong?
You can try this
c\.CreateMap\<\w+\.?\w+?\,\s*\w*?\.?Address\>
Explanation
c\.CreateMap\< - Matches c\.CreateMap\<.
\w+ - Matches any word character one or more time.
\.? - Matches '.' zero or one time.
\, - Matches ','.
\s* - Matches space zero or more time.
\w - Matches word character zero or more time.
\.? - Matches '.' zero or one time.
Address\> - Matches Address\>.
Demo
P.S- In case you also want to match something like this.
c.CreateMap<Responses.SiteAddress.abc, Data.Address.xyz>()
You can use this.
c\.CreateMap\<(\w+\.?\w+?)*\,\s*(?:\w*?\.?)*Address(\.\w*)?\>
Demo
Here is general regex I can suggest:
c\.CreateMap\<[\w.]+,\s+(?:[\w.]+\.)?Address\>\s*\(\s*\)
This will match any term with dots or word characters in the first position in the diamond. In the second, position, it will match Address, or some parent class names, followed by a dot separator, followed by Address.
Demo
Note that I also include the empty function call parentheses in the regex. As well, I allow for flexibility in the whitespace may appear after the diamond, or between the parentheses.
In your second example, you have extra dot which is not handled. Your regex needs little modification. Also, you don't need to escape < or > or , Use this,
c\.CreateMap<([\w.])+,\s+[\w.]*Address>
Demo
To match any of the functions on your question, you can use:
c\.CreateMap[^)]+\)
Regex Demo
Regex Explanation:

How to only match a single instance of a character?

Not quite sure how to go about this, but basically what I want to do is match a character, say a for example. In this case all of the following would not contain matches (i.e. I don't want to match them):
aa
aaa
fooaaxyz
Whereas the following would:
a (obviously)
fooaxyz (this would only match the letter a part)
My knowledge of RegEx is not great, so I am not even sure if this is possible. Basically what I want to do is match any single a that has any other non a character around it (except for the start and end of the string).
Basically what I want to do is match any single a that has any other non a character around it (except for the start and end of the string).
^[^\sa]*\Ka(?=[^\sa]*$)
DEMO
\K discards the previously matched characters and lookahead assertes whether a match is possibel or not. So the above matches only the letter a which satifies the conditions.
OR
a{2,}(*SKIP)(*F)|a
DEMO
You may use a combination of a lookbehind and a lookahead:
(?<!a)a(?!a)
See the regex demo and the regex graph:
Details
(?<!a) - a negative lookbehind that fails the match if, immediately to the left of the current location, there is a a char
a - an a char
(?!a) - a negative lookahead that fails the match if, immediately to the right of the current location, there is a a char.
You need two things:
a negated character class: [^a] (all except "a")
anchors (^ and $) to ensure that the limits of the string are reached (in other words, that the pattern matches the whole string and not only a substring):
Result:
^[^a]*a[^a]*$
Once you know there is only one "a", you can use the way you want to extract/replace/remove it depending of the language you use.

Regular expression let periods in (.)

My regular expression lets in periods for some reason, how can I keep that from happening.
Rules:
4-15 characters
Any alphanumeric characters
Underscore as long as it's not first or last
[A-Za-z][A-Za-z0-9_]{3,14}
I don't want "bad.example" for work.
Edit: changed to 4-15 characters
Your regex matches example as a substring of bad.example. Use anchors to prevent that:
^[A-Za-z][A-Za-z0-9_]{1,12}[A-Za-z]$
Note that (like your regex) this regex also prevents digits from matching in the first and last position - if they should be allowed (as per your specs), just add 0-9 at the end of the character classes.
^[A-Za-z][A-Za-z0-9_]{3,14}$
try this
This will match any alphanumeric at the beginning and end. In the middle it will accept from one up to twelve alphanumerics including an underscore:
^[a-zA-Z\d]\w{1,12}[a-zA-Z\d]$
It does not match bad.example but matches only example as your regex allows a character from 4 to 15.See here.
http://regex101.com/r/xV4eL5/5
To prevent it you need to match the whole input and not make partial matches.Put a ^ start anchor and $ end anchor.
Use
\A[A-Za-z0-9][\w]{1,12}[A-Za-z0-9]\Z

Regular expression to match non-integer values in a string

I want to match the following rules:
One dash is allowed at the start of a number.
Only values between 0 and 9 should be allowed.
I currently have the following regex pattern, I'm matching the inverse so that I can thrown an exception upon finding a match that doesn't follow the rules:
[^-0-9]
The downside to this pattern is that it works for all cases except a hyphen in the middle of the String will still pass. For example:
"-2304923" is allowed correctly but "9234-342" is also allowed and shouldn't be.
Please let me know what I can do to specify the first character as [^-0-9] and the rest as [^0-9]. Thanks!
This regex will work for you:
^-?\d+$
Explanation: start the string ^, then - but optional (?), the digit \d repeated few times (+), and string must finish here $.
You can do this:
(?:^|\s)(-?\d+)(?:["'\s]|$)
^^^^^ non capturing group for start of line or space
^^^^^ capture number
^^^^^^^^^ non capturing group for end of line, space or quote
See it work
This will capture all strings of numbers in a line with an optional hyphen in front.
-2304923" "9234-342" 1234 -1234
++++++++ captured
^^^^^^^^ NOT captured
++++ captured
+++++ captured
I don't understand how your pattern - [^-0-9] is matching those strings you are talking about. That pattern is just the opposite of what you want. You have simply negated the character class by using caret(^) at the beginning. So, this pattern would match anything except the hyphen and the digits.
Anyways, for your requirement, first you need to match one hyphen at the beginning. So, just keep it outside the character class. And then to match any number of digits later on, you can use [0-9]+ or \d+.
So, your pattern to match the required format should be:
-[0-9]+ // or -\d+
The above regex is used to find the pattern in some large string. If you want the entire string to match this pattern, then you can add anchors at the ends of the regex: -
^-[0-9]+$
For a regular expression like this, it's sometimes helpful to think of it in terms of two cases.
Is the first character messed up somehow?
If not, are any of the other characters messed up somehow?
Combine these with |
(^[^-0-9]|^.+?[^0-9])