I've tried googling this but all I can find is how to match until a known character occurs. In my case I don't know the character beforehand.
I know I can match the last character of a string with (.?)$, and I know that I can match until a character X occurs with (?:(?!X).)*, but how do I combine the two to match until the first occurence and not the matched occurence?
Examples:
character → char
test → t
no match → no match
This is a test → This is a t
I came. I saw. I conquered. → I came.
In pseudocode what I want is basically str.substring(0,str.indexOf(str.lastChar)).
You may use
^(?=(.)*$).*?\1
^(?=.*(.)$).*?\1
See the regex demo. If you need to match multiline strings, see how How do I match any character across multiple lines in a regular expression?.
Details
^ - start of string
(?=(.)*$) - a positive lookahead capturing each char other than line break chars up to the end of string (last one is saved in Group 1)
.*? - any 0 or more chars other than line break chars as few as possible
\1 - same char as in Group 1.
Related
there are 4 strings as shown below
ABC_FIXED_20220720_VALUEABC.csv
ABC_FIXED_20220720_VALUEABCQUERY_answer.csv
ABC_FIXED_20220720_VALUEDEF.csv
ABC_FIXED_20220720_VALUEDEFQUERY_answer.csv
Two strings are considered as matched based on a matching substring value (VALUEABC, VALUEDEF in the above shown strings). Thus I am looking to match first 2 (having VALUEABC) and then next 2 (having VALUEDEF). The matched strings are identified based on the same value returned for one regex group.
What I tried so far
ABC.*[0-9]{8}_(.*[^QUERY_answer])(?:QUERY_answer)?.csv
This returns regex group-1 (from (.*[^QUERY_answer])) value "VALUEABC" for first 2 strings and "VALUEDEF" for next 2 strings and thus desired matching achieved.
But the problem with above regex is that as soon as the value ends with any of the characters of "QUERY_answer", the regex doesn't match any value for the grouping. For instance, the below 2 strings doesn't match at all as the VALUESTU ends with "U" here :
ABC_FIXED_20220720_VALUESTU.csv
ABC_FIXED_20220720_VALUESTUQUERY_answer.csv
I tried to use Negative Lookahead:
ABC.*[0-9]{8}_(.*(?!QUERY_answer))(?:QUERY_answer)?.csv
but in this case the grouping-1 value is returned as "VALUESTU" for first string and "VALUESTUQUERY_answer" for second string, thus effectively making the 2 strings unmatched.
Any way to achieve the desired matching?
With your shown samples please try following regex.
^ABC_[^_]*_[0-9]+_(.*?)(?:QUERY_answer)?\.csv$
OR to match exact 8 digits try:
^ABC_[^_]*_[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv$
Here is the online demo for above regex.
Explanation: Adding detailed explanation for above regex.
^ABC_[^_]*_ ##Matching from starting of value ABC followed by _ till next occurrence of _.
[0-9]+_ ##Matching continuous occurrences of digits followed by _ here.
(.*?) ##Creating one and only capturing group using lazy match which is opposite of greedy match.
(?:QUERY_answer)? ##In a non-capturing group matching QUERY_answer and keeping it optional.
\.csv$ ##Matching dot literal csv at the end of the value.
You need
ABC.*[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv
See the regex demo.
Note
.*[^QUERY_answer] matches any zero or more chars other than line break chars as many as possible, and then any one char other than Q, U, E, etc., i.e. any char in the negated character class. This is replaced with .*?, to match any zero or more chars other than line break chars as few as possible.
(?:QUERY_answer)? - the group is made non-capturing to reduce grouping complexity.
\.csv - the . is escaped to match a literal dot.
I want to write a regex pattern to match a string starting with "Z" and not containing the next 2 characters as "IU" followed by any other characters.
I am using this pattern but it is not working Z[^(IU)]+.*$
ZISADR - should match
ZIUSADR - should not match
ZDDDDR - should match
Try this regex:
^Z(?:I[^U]|[^I]).*$
Click for Demo
Explanation:
^ - asserts the start of the line
Z - matches Z
I[^U] - matches I followed by any character that is not a U
| - OR
[^I] - matches any character that is not a I
.* - matches 0+ occurrences of any character that is not a new line
$ - asserts the end of the line
When you want to negate certain characters in a string, you can use character class but when you want to negate more than one character in a particular sequence, you need to use negative look ahead and write your regex like this,
^Z(?!IU).*$
Demo
Also note, your first word ZISADR will match as Z is not followed by IU
Your regex, Z[^(IU)]+.*$ will match the starting with Z and [^(IU)]+ character class will match any character other than ( I U and ) one or more times further followed by .* means it will match any characters zero or more times which is not the behavior you wanted.
Edit: To provide a solution without look ahead
A non-lookahead based solution would be to use this regex,
^Z(?:I[^U]|[^I]U|[^I][^U]).*$
This regex has three main alternations which incorporate all cases needed to cover.
I[^U] - Ensures if second character is I then third shouldn't be U
[^I]U - Ensures if third character is U then second shouldn't be I
[^I][^U] - Ensures that both second and third characters shouldn't be I and U altogether.
Demo non-look ahead based solution
I have three strings as list below:
Levofloxacin 500mg/100mL
Levofloxacin 500mg
Procaterol Hydrochloride …………… 25μg
The first line, I want to just get 'mg' without 'mL' in my result.
The second line, I want get 'mg'.
The third line, I want get 'ug'.
I have try regexp pattern like:
(?!(.*[ ]{1}[0-9]+))[a-zA-Zμ]+
However, the first line always returns 'mg' with 'mL'...
How could I just acquire 'mg' with regexp?
Any suggestions will be appreciated.
As mentioned in the comment section, try this regex:
^\D*[\d.]+\K[a-zμ]+
Click for Demo
Explanation:
^ - asserts the start of the string
\D* - matches 0+ occurrences of any character that is not a digit
[\d.]+ - matches 1+ occurrences of any character that is a digit
\K - removes what has been matched so far
[a-zμ]+ - this is what you want. This will contain the units like mg, ml appearing after the first number. If there are any other special characters like μ, you can add them too in this character list
I the below items i want to only detect the valid items with regular expression.
Space in word means invalid, # sign means invalid, Starting word with number is invalid.
Invalid : M_123 ASD
Invalid : M_123#ASD
Invalid : 1_M# ADD
Valid : M_125ASD
Valid : M_125$ASD
I am trying as below :
[A-Za-z0-9_$]
Not working properly. I need to set both valid and invalid sets for a word.
Can i do a match with regular expression?
Your regex [A-Za-z0-9_$] presents a character class that matches a single character that is either an ASCII letter or digit, or _ or $ symbols. If you use it with std::regex_match, it would only match a whole string that consists of just one char like that since the pattern is anchored by default when used with that method. If you use it with an std::regex_search, a string like ([_]) would pass, since the regex is not anchored and can find partial matches.
To match 0 or more chars, you need to add * quantifier after your class. To match one or more chars, you need to add + quantifier after your character class. However, you have an additional restriction: a digit cannot appear at the start.
It seems you may use
^[A-Za-z][A-Za-z0-9_$]*$
See the regex demo at regex101.com.
Details:
^ - start of string
[A-Za-z] - an ASCII letter (exactly one occurrence)
[A-Za-z0-9_$]* - 0+ ASCII letters, digits, _ or $
$ - end of string anchor.
Note that with regex_match, you may omit ^ and $ anchors.
So the requirements are
cannot start with number( i am assuming it as start with alphabet)
cannot contain space or #
all other characters are valid
you can try this regex ^[a-zA-Z]((?![\# ]).)+?$
^[a-zA-Z] checks for alphabet at start of the line
((?![\# ]).)+?$ checks if there are no # or space in the remaining part of the line.
Online demo here
EDIT
As per Wiktor's comment the regex can be simplified to ^[a-zA-Z][^# ]+$.
I need to find into multiple strings two words with no words or only one word between them. I created the regex for the case to find if those two words exist in string:
^(?=[\s\S]*\bFirst\b)(?=[\s\S]*\bSecond\b)[\s\S]+
and it works correctly.
Then I tried to insert in this regex additional code:
^(?=[\s\S]*\bFirst\b)(\b\w+\b){0,1}(?=[\s\S]*\bSecond\b)[\s\S]+
but it didn't work. It selects text with two or more words between searched words. It is not what I need.
First Second - must be selected
First word1 Second - must be selected
First word1 word2 Second - must be not selected by regex, but my regex select it.
Can I get advise how to solve this problem?
Root cause
You should bear in mind that lookarounds match strings without moving along the string, they "stand their ground". Once you write ^(?=[\s\S]*\bFirst\b)(\b\w+\b){0,1}(?=[\s\S]*\bSecond\b), the execution is as follows:
^ - the regex engine checks if the current position is the start of string
(?=[\s\S]*\bFirst\b) - the positive lookahead requires the presence of any 0+ chars followed with a whole word First - note that the regex index is still at the start of the string after the lookahead returns true or false
(\b\w+\b){0,1} - this subpattern is checked only if the above check was true (i.e. there is a whole word First somewhere) and matches (consumes, moves the regex index) 1 or 0 occurrences of a whole word (i.e. there must be 1 or more word chars right at the string start
(?=[\s\S]*\bSecond\b) - another positive lookahead that makes sure there is a whole word Second somewhere after the first whole word consumed with \b\w+\b - if any. Even if the word Second is the first word in the string, this will return true since backtracking will step back the word matched with (\b\w+\b){0,1} (see, it is optional), and the Second will get asserted, and [\s\S]+ will grab the whole string (Group 1 will be empty). See the regex demo with Second word word2 First string.
So, your approach cannot guarantee the order of First and Second in the string, they are just required to be present but not necessarily in the order you expect.
Solution
If you need to check the order of First and Second in the string, you need to combine all the checks into one single lookahead. The approach might turn out very inefficient with longer strings and multiple alternatives in the lookaround, consider either unrolling the patterns, or trying mutliple regex patterns (like this pseudo-code if /\bFirst\b/.finds_match().index < /\bSecond\b/.finds_match().index => Good, go on...).
If you plan to go on with the regex approach, you may match a string that contains First....Second only in this order:
^(?=[\s\S]*\bFirst(?:\W+\w+)?\W+Second\b)[\s\S]+
See the regex demo
Details:
^ - start of string
(?=[\s\S]*\bFirst(?:\W+\w+)?\W+Second\b) - there must be:
[\s\S]* - any zero or more chars up to the last
\bFirst - whole word First
(?:\W+\w+)? - optional sequence (1 or 0 occurrences) of 1+ non-word chars and 1+ word chars
\W+ - 1+ non-word chars
Second\b - Second as a whole word
[\s\S]+ - any 1 or more characters (empty string won't match).