I'm using Regex to match whole sentences in a text containing a certain string. This is working fine as long as the sentence ends with any kind of punctuation. It does not work however when the sentence is at the end of the text without any punctuation.
This is my current expression:
[^.?!]*(?<=[.?\s!])string(?=[\s.?!])[^.?!]*[.?!]
Works for:
This is a sentence with string. More text.
Does not work for:
More text. This is a sentence with string
Is there any way to make this word as intended? I can't find any character class for "end of text".
End of text is matched by the anchor $, not a character class.
You have two separate issues you need to address: (1) the sentence ending directly after string, and (2) the sentence ending sometime after string but with no end-of-sentence punctuation.
To do this, you need to make the match after string optional, but anchor that match to the end of the string. This also means that, after you recognize an (optional) end-of-sentence punctuation mark, you need to match everything that follows, so the end-of-string anchor will match.
My changes: Take everything after string in your original regex and surround it in (?:...)? - the (?:...) being a "non-remembered" group, and the ? making the entire group optional. Follow that with $ to anchor the end of the string.
Within that optional group, you also need to make the end-of-sentence itself optional, by replacing the simple [.?!] with (?:[.?!].*)? - again, the (?:...) is to make a "non-remembered" group, the ? makes the group optional - and the .* allows this to match as much as you want after the end-of-sentence has been found.
[^.?!]*(?<=[.?\s!])string(?:(?=[\s.?!])[^.?!]*(?:[.?!].*)?)?$
The symbol for end-of-text is $ (and, the symbol for beginning-of-text, if you ever need it, is ^).
You probably won't get what you're looking for with by just adding the $ to your punctuation list though (e.g., [.?!$]); you'll find it works better as an alternative choice: ([.?!]|$).
Your regex is way too complex for what you want to achieve.
To match only a word just use
"\bstring\b"
It will match start, end and any non-alphanum delimiters.
It works with the following:
string is at the start
this is the end string
this is a string.
stringing won't match (you don't want a match here)
You should add the language in the question for more information about using.
Here is my example using javascript:
var reg = /^([\w\s\.]*)string([\w\s\.]*)$/;
console.log(reg.test('This is a sentence with string. More text.'));
console.log(reg.test('More text. This is a sentence with string'));
console.log(reg.test('string'))
Note:
* : Match zero or more times.
? : Match zero or one time.
+ : Match one or more times.
You can change * with ? or + if you want more definition.
Related
I am new to regex, basically I'd like to check if a word has ONLY one colons or not.
If has two or more colons, it will return nothing.
if has one colon, then return as it is. (colon must be in the middle of string, not end or beginning.
(1)
a:bc:de #return nothing or error.
a:bc #return a:bc
a.b_c-12/:a.b_c-12/ #return a.b_c-12/:a.b_c-12/
(2)
My thinking is, but this is seems too complicated.
^[^:]*(\:[^:]*){1}$
^[-\w.\/]*:[-\w\/.]* #this will not throw error when there are 2 colons.
Any directions would be helpful, thank you!
This will find such "words" within a larger sentence:
(?<= |^)[^ :]+:[^ :]+(?= |$)
See live demo.
If you just want to test the whole input:
^[^ :]+:[^ :]+$
To restrict to only alphanumeric, underscore, dashes, dots, and slashes:
^[\w./-]+:[\w./-]+$
I saw this as a good opportunity to brush up on my regex skills - so might not be optimal but it is shorter than your last solution.
This is the regex pattern: /^[^:]*:[^:]*$/gm and these are the strings I am testing against: 'oneco:on' (match) and 'one:co:on', 'oneco:on:', ':oneco:on' (these should all not match)
To explain what is going on, the ^ matches the beginning of the string, the $ matches the end of the string.
The [^:] bit says that any character that is not a colon will be matched.
In summary, ^[^:] means that the first character of the string can be anything except for a colon, *: means that any number of characters can come after and be followed by a single colon. Lastly, [^:]*$ means that any number (*) of characters can follow the colon as long as they are not a colon.
To elaborate, it is because we specify the pattern to look for at the beginning and end of the string, surrounding the single colon we are looking for that only the first string 'oneco:on' is a match.
I have a comma separated string which I want to validate using a regex. What I have written is gives me a match if there a part wrong later in the string. I want to discard it completely if any part is wrong.
My regex : ^(?:[\w\.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+
Positive Case : Component,STR,YoungGenUse,STR,YoungGenMax,STR,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
Negative Case :
Component,STR,YoungGenUse,STR,YoungGenMax,TEST,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
For the second case, my regex gives a match for the bold portion eventhough, later there is an incorrect part (TEST). How can I modify my regex to discard the entire string?
The pattern that you tried would not match TEST in YoungGenMax,TEST because the alternatives STR|INT|REAL do not match it.
It would show until the last successful match in the repetition which would be Component,STR,YoungGenUse,STR,
You have to add the anchor at the end, outside of the repetition of the group, to indicate that the whole pattern should be followed by asserting the end of the string.
There are no spaces or dots in your string, so you might leave out \s* and use \w+ without the dot in the character class. Note that \s could also possibly match a newline.
^(?:\w+,(?:STR|INT|REAL),)+$
Regex demo
If you want to keep matching optional whitespace chars and the dot:
^(?:[\w.]+,(?:STR|INT|REAL),\s*)+$
Regex demo
Note that by repeating the group with the comma at the end, the string should always end with a comma. You can omit {1} from the pattern as it is superfluous.
your regex must keep matching until end of the string, so you must use $ to indicate end of the line:
^(?:[\w.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+$
Regex Demo
I'd like to find word RADU3_ or RADU3- in a sentence that begins with xlink:href= and ends with .svg
How to do this?
I've tried following, but does not give the result I'm expecting.
(?=\wxlink:href=|\wsvg\b)|\bRADU3_|\bRADU3-
Just last line in example is good result (RADU3_)
ProductionGraphics\GP1**RADU3-**11_HeatingFurnaceF1.svg
PB:ExpressionText id="RADU3_FUEL GAS _SUM_EX" PBD:LinkUses
xlink:href="C:\ProcBookImport\MaintenanceGraphics\RADU3_AI.svg"
Example...
Not sure exactly how you want to use it but the below pattern finds the string. I put the RADU3 part in a group where I matches RADU3 followed by - or _ ([_-])
(xlink:href=.*)(RADU3[_-]*)(.*\.svg)
Edit, handle multiple occurences
If a string might contain the pattern several times then use ? to allow a group to repeat itself
(RADU3[_-]*?)(.*?\.svg?)
The above could be used in a replace expression like
\1someotherword\3
Where \2 is the second group that is replaced
If you want to make sure that the string starts with xlink:href= and ends with \.svg you could use anchors to assert the start ^ and the end $ of the string.
Use 1 capturing group to make sure xlink:href= comes before RADU3 followed by an underscore or a hyphen. Then you could match it and in the replacement use that capturing group follwed by your replacement.
You could use a positive lookahead to assert that the string ends with \.svg
That will match:
^(xlink:href=.*)\bRADU3[_-](?=.*\.svg$)
^ Assert the start of the string
(xlink:href=.*) Capturing group, match up until the last occurence of ..
\bRADU3[_-] Word boundary to prevent matching part of a larger word. Match RADU3 followed by an underscore or hyphen
(?=.*\.svg$) Positive lookahead to assert the string ends with .svg
See the regex demo
It sounds like you only want the word (substring) if it is in a specific context?
In your case, you can restart the regex midways if you want to have starting and ending conditions (multiple conditions) for a string, but at the same time only want to use these conditions as "if-statements" and not as part of the result.
The following uses this method, and utilizes restarts (\K) in order to only extract the substring you are looking for.
# The string has to start with "xlink:href="
xlink:href=
# Fetch everything up to our match, and the restart the regex
.*\K
# The strings we are looking for
(RADU3[-_])
# String has to end with ".svg"
(?=(.*\.svg))
If you want the entire string matching our rules you are looking for something like this:
#The string has to start with "xlink:href"
^(xlink:href=).*
# The strings we are looking for
(RADU3[-_])
# String has to end with ".svg"
(\w+\.svg)
#Get everything after .svg too
.*
If you only want the ending " after the .svg, you'd want to modify the last part where I just take everything after .svg
You can play around with what I have come up with at regex101 (no affiliation, just love their site): https://regex101.com/r/g0v07V/3/
My question is pretty similar to this question and the answer is almost fine. Only I need a regexp not only for character-to-character but for a second occurance of a character till a character.
My purpose is to get password from uri, example:
http://mylogin:mypassword#mywebpage.com
So in fact I need space from the second ":" till "#".
You could give the following regex a go:
(?<=:)[^:]+?(?=#)
It matches any consecutive string not containing any : character, prefixed by a : and suffixed by a #.
Depending on your flavour of regex you might need something like:
:([^:]+?)#
Which doesn't use lookarounds, this includes the : and # in the match, but the password will be in the first capturing group.
The ? makes it lazy in case there should be any # characters in the actual url string, and as such it is optional. Please note that that this will match any character between : and # even newlines and so on.
Here's an easy one that does not need look-aheads or look-behinds:
.*:.*:([^#]+)#
Explanation:
.*:.*: matches everything up to (and including) the second colon (:)
([^#]+) matches the longest possible series of non-# characters
# - matches the # character.
If you run this regex, the first capturing group (the expression between parentheses) will contain the password.
Here it is in action: http://regex101.com/r/fT6rI0
I've been working for many hours trying to do a "simple thing": use a regex to validate a text field.
I need to make sure of:
1- Only use (a-z), (A-Z) and (0-9) values
2- Add a SINGLE wildcard only at the end.
Ex.
Match
MICHE*
Match
JAMES
No match
MICHE**
No match
MIC_HEAL*
I have this regex till now:
[a-zA-Z0-9\s-]+.\z*?
The problem is it still matches when I introduce an invalid character as long as I have a matching sub-string See my REGEX
What can I do to force a match on the whole string? What am I missing?
Thx!
Use ^ (start of line) and $ (end of line) to only match the whole string:
^[a-zA-Z0-9\s-]+.\z*?$
(If you have a multiline input you can also use \A and \z - start and end of string)
On a second look, I don't understand the end of your regex: . (anything) \z * ? (end of string, zero or more times, zero or one time). This regex will match something like:
Ikdflfdf&
Is that correct? If you only want the character *, you should use:
^[a-zA-Z0-9\s-]+\*?$
Also, as Robbie pointed out, you're including spaces and the - in your list of accepted characters. If you only want letters and digits, a shortcut would be using \w (word characters):
^\w+\*$
However, depending on whether the matcher is Unicode-aware or not, \w will also match non-ASCII letters and digits, which may or may not be what you want.
Try this one :
^[a-zA-Z0-9]+\*?$
^ string start
$ string end
* is meta character so it should be escaped like \* to use it as a letter
I think you just need ^ at the begining and $ at the end
^[a-zA-Z0-9\s-]+.\*?$
Also, you don't need the \z
Also, you haven't mentioned that you want to allow spaces and dashes - but you have included them in your allowed character set.