RegEx in JS to find No 3 Identical consecutive characters - javascript

How to find a sequence of 3 characters, 'abb' is valid while 'abbb' is not valid, in JS using Regex (could be alphabets,numerics and non alpha numerics).
This question is a variation of the question that I have asked in here : How to combine these regex for javascript.
This is wrong : /(^([0-9a-zA-Z]|[^0-9a-zA-Z]))\1\1/ , so what is the right way to do it?

This depends on what you actually mean. If you only want to match three non-identical characters (that is, if abb is valid for you), you can use this negative lookahead:
(?!(.)\1\1).{3}
It first asserts, that the current position is not followed by three times the same character. Then it matches those three characters.
If you really want to match 3 different characters (only stuff like abc), it gets a bit more complicated. Use these two negative lookaheads instead:
(.)(?!\1)(.)(?!\1|\2).
First match one character. Then we assert, the this is not followed by the same character. If so, we match another character. Then we assert that these are followed neither by the first nor the second character. Then we match a third character.
Note that those negative lookaheads ((?!...)) do not consume any characters. That is why they are called lookaheads. They just check what is coming next (or in this case what is not coming next) and then the regex continues from where it left of. Here is a good tutorial.
Note also that this matches anything but line breaks, or really anything if you use the DOTALL or SINGLELINE option. Since you are using JavaScript you can just activate the option by appending s after the regexes closing delimiter. If (for some reason) you don't want to use this option, replace the .s by [\s\S] (this always matches any character).
Update:
After clarification in the comments, I realised that you do not want to find three non-identical characters, but instead you want to assert that your string does not contain three identical (and consecutive) characters.
This is a bit easier, and closer to your former question, since it only requires one negative lookahead. What we do is this: we search the string from the beginning for three consecutive identical characters. But since we want to assert that these do not exist we wrap this in a negative lookahead:
^(?!.*(.)\1\1)
The lookahead is anchored to the beginning of the string, so this is the only place where we will look. The pattern in the lookahead then tries to find three identical characters from any position in the string (because of the .*; the identical characters are matched in the same way as in your previous question). If the pattern finds these, the negative lookahead will thus fail, and so the string will be invalid. If not three identical characters can be found, the inner pattern will never match, so the negative lookahead will succeed.

To find non-three-identical characters use regex pattern
([\s\S])(?!\1\1)[\s\S]{2}

Related

How to match one 'x' but not one or both of xs in 'xx' globally in string [duplicate]

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.

how to negate a capture group?

Using a javascript regexp, I would like to find strings like "/foo" or "/foo d/" but not "/foo /"; ie, "annotation character", then either word with no terminating annotation, or multiple words, where the termination comes at the end of the phrase (with no space). Complicating the situation, there are three possible annotation symbols: /, \ and |.
I've tried something like:
/(?:^|\s)([\\\/|])((?:[\w_-]+(?![^\1]+[\w_-]\1))|(?:[\w\s]+[\w](?=\1)))/g
That is, start with space, then annotation, then
word not followed by (anything but annotation) then letter and annotation... or
possibly multiple words, immediately followed by annotation character.
The problem is the [^\1]: this doesn't read as "anything but the annotation character" in the angle brackets.
I could repeat the whole phrase three times, one for each annotation character. Any better ideas?
As you've mentioned, [^\1] doesn't work - it matches anything that is not the character 1. In JavaScript, you can negate \1 by using a lookahead: (?:(?!\1).)* . This is not as efficient, but it works.
Your pattern can be written as:
([\\\/|])([\w\-]+(?:(?:(?!\1).)*[\w\-]\1)?)
Working example at Regex101
\w already contains underscore.
Instead of alternation (a|ab) I'm using an optional group (a(?:b)?) - we always match the first word, with optional further words and tags.
You may still want to include (?:^|\s) at the beginning.

Regexp: excluding a word but including non-standard punctuation

I want to find strings that contain words in a particular order, allowing non-standard characters in between the words but excluding a particular word or symbol.
I'm using javascript's replace function to find all instances and put into an array.
So, I want select...from, with anything except 'from' in between the words. Or I can separate select...from from select...from (, as long as I exclude nesting. I think the answer is the same for both, i.e. how do I write: find x and not y within the same regexp?
From the internet, I feel this should work: /\bselect\b^(?!from).*\bfrom\b/gi but this finds no matches.
This works to find all select...from: /\bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b/gi but modifying it to exclude the parenthesis "(" at the end prevents any matches: /\bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b\s*^\(/gi
Can anyone tell me how to exclude words and symbols within this regexp?
Many thanks
Emma
Edit: partial string input:
left outer join [stage].[db].[table14] o on p.Project_id = o.project_id
left outer join
(
select
different_id
,sum(costs) - ( sum(brushes) + sum(carpets) + sum(fabric) + sum(other) + sum(chairs)+ sum(apples) ) as overallNumber
from
(
select ace from [stage].db.[table18] J
Javascript:
sequel = stringInputAsAbove;
var tst = sequel.replace(/\bselect\b[\s\S]*?\bfrom\b/gi, function(a,b) { console.log('match: '+a); selects.push(b); return a; });
console.log(selects);
Console.log(selects) should print an array of numbers, where each number is the starting character of a select...from. This works for the second regexp I gave in my info, printing: [95, 251]. Your \s\S variation does the same, #stribizhev.
The first example ^(?!from).* should do likewise but returns [].
The third example \s*^\( should return 251 only but returns []. However I have just noticed that the positive expression \s*\( does give 95, so some progress! It's the negatives I'm getting wrong.
Your \bselect\b^(?!from).*\bfrom\b regex doesn't work as expected because:
^ means here beginning of a line, not negation of next part, so
the \bselect\b^ means, select word followed by beginning of a
line. After removal of ^ regex start to match something
(DEMO) but it is still invalid.
in multiline text .* without modification will not match new line,
so regex will match only select...from in single lines, but if you
change it for (.|\n)* (as a simple example) it will match
multiline, but still invalid
the * is greede quantifire, so it will match as much a possible,
but if you use reluctant quantifire *?, regex will match to first
occurance of from word, and int will start to return relativly
correct result.
\bselect\b(?!from) means match separate select word which is not
directly followed by separate from word, so it would be
selectfrom somehow composed of separate words (because
select\bfrom) so (?!from) doesn't work and it is redundant
In effect you will get regex very similar to what Stribizhev gave you: \bselect\b(.|\n)*?\bfrom\b
In third expression you meke same mistake: \bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b\s*^\( using ^ as (I assume) a negation, not beginning of a line. Remove ^ and you will again get relativly valid result (match from select through from to closing parathesis ) ).
Your second regex works similar to \bselect\b(.|\n)*?\bfrom\b or \bselect\b[\s\S]*?\bfrom\b.
I wrote "relativly valid result", as I also think, that parsing SQL with regex could be very camplicated, so I am not sure if it will work in every case.
You can also try to use positive lookahead to match just position in text, like:
(?=\bselect\b(?:.|\n)*?\bfrom\b)
DEMO - the () was added to regex just to return beginning index of match in groups, so it would be easier to check it validity
Negation in regex
We use ^ as negation in character class, for example [^a-z] means match anything but not letter, so it will match number, symbol, whitespace, etc, but not letter from range a to z (Look here). But this negation is on a level of single character. I you use [^from] it will prevent regex from matching characters f,r,o and m (demo). Also the [^from]{4} will avoid matching from but also form, morf, etc.
To exlude the whole word from matching by regex, you need to use negative look ahead, like (?!from), which will fail to match, if there will be chosen word from fallowing given position. To avoid matching whole line containing from you could use ^(?!.*from.*).+$ (demo).
However in your case, you don't need to use this construction, because if you replace greedy quantifire .*\bfrom with .*?\bfrom it will match to first occurance of this word. Whats more it would couse problems. Take a look on this regex, it will not match anything because (?![\s\S]*from[\s\S]*) is not restricted by anything, so it will match only if there is no from after select, but we want to match also from! in effect this regex try to match and exclude from at once, and fail. so the (?!.*word.*) construction works much better to exclude matching line with given word.
So what to do if we don't what to match a word in a fragment of a match? I think select\b([^f]|f(?!rom))*?\bfrom\b is a good solution. With ([^f]|f(?!rom))*? it will match everything between select and from, but will not exclude from.
But if you would like to match only select...from not followed by ( then it is good idea to use (?!\() like. But in your regex (multiline, use of (.|\n)*? or [\s\S]*? it will cause to match up to next select...from part, because reluctant quantifire will chenge a plece where it need to match to make whole regex . In my opinion, good solution would be to use again:
select\b([^f]|f(?!rom))*?\bfrom\b(?!\s*?\()
which will not overlap additional select..from and will not match if there is \( after select...from - check it here

The behavior of /g mode matching

On this article, it mentioned
Make sure you are clear on the fact that an expression pattern is
tested on each individual character. And that, just because the engine
moves forward when following the pattern and looking for a match it
still backtracks and examines each character in a string until a match
is found or if the global flag is set until all characters are
examined.
But what I tested in Javascript
"aaa#bbb".match(/a+#b+/g)
does not produce a result like:
["aaa#bbb", "aa#bbb", "a#bbb"]
It only produces ["aaa#bbb"]. It seems it does not examine each character to test the pattern. Can anyone can explain a little on matching steps ? Thanks.
/g does not mean it will try to find every possible subset of characters in the input string which may match the given pattern. It means that once a match is found, it will continue searching for more substrings which may match the pattern starting from the previous match onward.
For example:
"aaa#bbb ... aaaa#bbbb".match(/a+#b+/g);
Will produce
["aaa#bbb", "aaaa#bbbb"]
That explanation is mixing two distinct concepts that IMO should be kept separated
A) backtracking
When looking for a match the normal behavior for a quantifier (?, *, +) is to be "greedy", i.e. to munch as much as possible... for example in /(a+)([^b]+)/ tested with aaaacccc all the a will be part of group 1 even if of course they also match the char set [^b] (everything but b).
However if grabbing too much is going to prevent a match the RE rules require that the quantifier "backtracks" capturing less if this allows the expression to match. For example in (a+)([^b]+) tested with aaaa the group 1 will get only three as, to leave one for group 2 to match.
You can change this greedy behavior with "non-greedy quantifiers" like *?, +?, ??. In this case stills the engine does backtracking but with the reverse meaning: a non-greedy subexpression will eat as little as possible that allows the rest of expression to match. For example (a+)(a+b+) tested with aaabbb will leave two as for group 1 and abbb for group 2, but (a+?)(a+b+) with the same string instead will leave only one a for group 1 because this is the minimum that allows matching the remaining part.
Note that also because of backtracking the greedy/non-greedy options doesn't change if an expression has a match or not, but only how big is the match and how much goes to each sub-expression.
B) the "global" option
This is something totally unrelated to backtracking and simply means that instead of stopping at the first match the search must find all non-overlapping matches. This is done by finding the first match and then starting again the search after the end of the match.
Note that each match is computed using the standard regexp rules and there is no look-ahead or backtracking between different matches: in other words if making for example a greedy match shorter would give more matches in the string this option is not considered... a+[^b]+ tested with aaaaaa is going to give only one match even if g option is specified and even if the substrings aa, aa, aa would have been each a valid match for the regexp.
When the global flag is used, it starts searching for the next match after the end of the previous match, to prevent generating lots of overlapping matches like that.
If you don't specify /g, the engine will stop as soon as a match is found.
If you do specify /g, it will keep going after a match. It will, however, still not produce overlapping matches, which is what you're asking about.
Its because.,
What Regex try to do:
All regex expression will try to match the best match.
What Regex wont do
It will not match the combinations for a single match as in your case.
When your "aaa#bbb".match(/a+#b+/g) scenario works
Rather, aaa#bbbHiaa#bbbHelloa#bbbSEEYOU try for some thing like this, which will give you
aaa#bbb
aa#bbb
a#bbb

Please explain some Javascript Regular Expressions

I'm learning Javascript via an online tutorial, but nowhere on that website or any other I googled for was the jumble of symbols explained that makes up a regular expression.
Check if all numbers: /^[0-9]+$/
Check if all letters: /^[a-zA-Z]+$/
And the hardest one:
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
What do all the slashes and dollar signs and brackets mean? Please explain.
(By the way, what languages are required to create a flexible website? I know a bit of Javascript and wanna learn jQuery and PHP. Anything else needed?)
Thanks.
There are already a number of good sites that explain regular expressions so I'll just dive a bit into how each of the specific examples you gave translate.
Check if all numbers: ^ anchors the start of the expression (e.g. start at the beginning of the text). Without it a match could be found anywhere. [0-9] finds the characters in that character class (e.g. the numbers 0-9). The + after the character class just means "one or more". The ending $ anchors the end of the text (e.g. the match should run to the end of the input). So if you put that together, that regular expression would allow for only 1 or more numbers in a string. Note that the anchors are important as without them it might match something like "foo123bar".
Check if all letters: Pretty much the same as above but the character classes are different. In this example the character class [a-zA-Z] represents all lowercase and uppercase characters.
The last one actually isn't any more difficult than the other two it's just longer. This answer is getting quite long so I'll just explain the new symbols. A \w in a character class will match word characters (which are defined per regex implementation but are generally 0-9a-zA-Z_ at least). The backslash before the # escapes the # so that it isn't seen as a token in the regex. A period will match any character so .+ will match one or more of any character (e.g. a, 1, Z, 1a, etc). The last part of the regex ({2,4}) defines an interval expression. This means that it can match a minimum of 2 of the thing that precedes it, and a maximum of 4.
Hope you got something out of the above.
There is an awesome explanation of regular expressions at http://www.regular-expressions.info/ including notes on language and implementation specifics.
Let me explain:
Check if all numbers: /^[0-9]+$/
So, first thing we see is the "/" at the beginning and the end. This is a deliminator, and only serves to show the beginning and end of the regular expression.
Next, we have a "^", this means the beginning of the string. [0-9] means a number from 0-9. + is a modifier, which modifies the term in front of it, in this case, it means you can have one or more of something, so you can have one or more numbers from 0-9.
Finally, we end with "$", which is the opposite of "^", and means the end of the string. So put that all together and it basically makes sure that inbetween the start and end of the string, there can be any number of digits from 0-9.
Check if all letters: /^[a-zA-Z]+$/
We notice this is very similar, but instead of checking for numbers 0-9, it checks for letters a-z (lowercase) and A-Z (uppercase).
And the hardest one:
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
"\w" means that it is a word, in this case we can have any number of letters or numbers, as well as the period means that it can be pretty much any character.
The new thing here is escape characters. Many symbols cannot be used without escaping them by placing a slash in front, as is the case with "\#". This means it is looking directly for the symbol "#".
Now it looks for letters and symbols, a period (this one seems incorrect, it should be escaping the period too, though it will still work, since an unescaped period will make any symbol). Numbers inside {} mean that there is inbetween this many terms in the previous term, so of the [a-zA-Z0-9], there should be 2-4 characters (this part here is the website domain, such as .com, .ca, or .info). Note there's another error in this one here, the [a-zA-z0-9] should be [a-zA-Z0-9] (capital Z).
Oh, and check out that site listed above, it is a great set of tutorials too.
Regular Expressions is a complex beast and, as already pointed out, there are quite a few guides off of google you can go read.
To answer the OP questions:
Check if all numbers: /^[0-9]+$/
regexps here are all delimated with //, much like strings are quoted with '' or "".
^ means start of string or line (depending on what options you have about multiline matching)
[...] are called character classes. Anything in [] is a list of single matching characters at that position in this case 0-9. The minus sign has a special meaning of "sequence of characters between". So [0-9] means "one of 0123456789".
+ means "1 or more" of the preceeding match (in this case [0-9]) so one or more numbers
$ means end of string/line match.
So in summary find any string that contains only numbers, i.e '0123a' will not match as [0-9]+ fails to match a before $).
Check if all letters: /^[a-zA-Z]+$/
Hopefully [A-Za-z] makes sense now (A-Z = ABCDEF...XYZ and a-z abcdef...xyz)
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
Not all regexp parses know the \w sequence. Javascript, java and perl I know do support it.
I have already have covered '/^ at the beginning, for this [] match we are looking for
\w - . and +. I think that regexp is incorrect. Either the minus sign should be escaped with \ or it should be at the end of the [] (i.e [\w+.-]). But that is an aside they are basically attempting to allow anything of abcdefghijklmnopqrstuvwxyz01234567890-.+
so fred.smith-foo+wee#mymail.com will match but fred.smith%foo+wee#mymail.com wont (the % is not matched by [\w.+-]).
\# is the litteral atsil sign (it is escaped as perl expands # an array variable reference)
[a-zA-Z0-9.-]+ is the same as [\w.-]+. Very much like the user part of the match, but does not match +. So this matches foo.com. and google.co. but not my+foo.com or my***domain.co.
. means match any one character. This again is incorrect as fred#foo%com will match as . matches %*^%$£! etc. This should of been written as \.
The last character class [a-zA-z0-9]{2,4} looks for between 2 3 or 4 of the a-zA-Z0-9 specified in the character class (much like + looks for "1 more more" {2,4} means at least 2 with a maximum of 4 of the preceeding match. So 'foo' matches, '11' matches, '11111' does not match and 'information' does not.
The "tweaked" regexp should be:
/^[\w.+-]+\#[a-zA-Z0-9.-]+\.[a-zA-z0-9]{2,4}$/
I'm not doing a tutorial on RegEx's, that's been done really well already, but here are what your expressions mean.
/^<something>$/ String begins, has something in the middle, and then immediately ends.
/^foo$/.test('foo'); // true
/^foo$/.test('fool'); // false
/^foo$/.test('afoo'); // false
+ One or more of something:
/a+/.test('cot');//false
/a+/.test('cat');//true
/a+/.test('caaaaaaaaaaaat');//true
[<something>] Include any characters found between the brackets. (includes ranges like 0-9, a-z, and A-Z, as well as special codes like \w for 0-9a-zA-Z_-
/^[0-9]+/.test('f00')//false
/^[0-9]+/.test('000')//true
{x,y} between X and Y occurrences
/^[0-9]{1,2}$/.test('12');// true
/^[0-9]{1,2}$/.test('1');// true
/^[0-9]{1,2}$/.test('d');// false
/^[0-9]{1,2}$/.test('124');// false
So, that should cover everything, but for good measure:
/^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
Begins with at least character from \w, -, +, or .. Followed by an #, followed by at least one in the set a-zA-Z0-9.- followed by one character of anything (. means anything, they meant \.), followed by 2-4 characters of a-zA-z0-9
As a side note, this regular expression to check emails is not only dated, but it is very, very, very incorrect.

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