Develop Reference

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.

Filter a string with regular expressions in javascript [duplicate]

How do I create a regular expression to match a word at the beginning of a string?
We are looking to match stop at the beginning of a string and anything can follow it.
For example, the expression should match:
stop
stop random
stopping

If you wish to match only lines beginning with stop, use
^stop
If you wish to match lines beginning with the word stop followed by a space:
^stop\s
Or, if you wish to match lines beginning with the word stop, but followed by either a space or any other non-word character you can use (your regex flavor permitting)
^stop\W
On the other hand, what follows matches a word at the beginning of a string on most regex flavors (in these flavors \w matches the opposite of \W)
^\w
If your flavor does not have the \w shortcut, you can use
^[a-zA-Z0-9]+
Be wary that this second idiom will only match letters and numbers, no symbol whatsoever.
Check your regex flavor manual to know what shortcuts are allowed and what exactly do they match (and how do they deal with Unicode).

Try this:
/^stop.*$/
Explanation:
/ charachters delimit the regular expression (i.e. they are not part of the Regex per se)
^ means match at the beginning of the line
. followed by * means match any character (.), any number of times (*)
$ means to the end of the line
If you would like to enforce that stop be followed by a whitespace, you could modify the RegEx like so:
/^stop\s+.*$/
\s means any whitespace character
+ following the \s means there has to be at least one whitespace character following after the stop word
Note: Also keep in mind that the RegEx above requires that the stop word be followed by a space! So it wouldn't match a line that only contains: stop

If you want to match anything after a word, stop, and not only at the start of the line, you may use: \bstop.*\b - word followed by a line.
Or if you want to match the word in the string, use \bstop[a-zA-Z]* - only the words starting with stop.
Or the start of lines with stop - ^stop[a-zA-Z]* for the word only - first word only.
The whole line ^stop.* - first line of the string only.
And if you want to match every string starting with stop, including newlines, use: /^stop.*/s - multiline string starting with stop.

Like #SharadHolani said. This won't match every word beginning with "stop"
. Only if it's at the beginning of a line like "stop going".
#Waxo gave the right answer:
This one is slightly better, if you want to match any word beginning with "stop" and containing nothing but letters from A to Z.
\bstop[a-zA-Z]*\b
This would match all
stop (1)
stop random (2)
stopping (3)
want to stop (4)
please stop (5)
But
/^stop[a-zA-Z]*/
would only match (1) until (3), but not (4) & (5)

If you want to match anything that starts with "stop" including "stop going", "stop" and "stopping" use:
^stop
If you want to match the word stop followed by anything as in "stop going", "stop this", but not "stopped" and not "stopping" use:
^stop\W

/stop([a-zA-Z])+/
Will match any stop word (stop, stopped, stopping, etc)
However, if you just want to match "stop" at the start of a string
/^stop/
will do :D

If you want the word to start with "stop", you can use the following pattern.
"^stop.*"
This will match words starting with stop followed by anything.

/^stop*$/i
i - in case it is case sensitive.

I'd advise against a simple regular expression approach to this problem. There are too many words that are substrings of other unrelated words, and you'll probably drive yourself crazy trying to overadapt the simpler solutions already provided.
You'll want at least a naive stemming algorithm (try the Porter stemmer; there's available, free code in most languages) to process text first. Keep this processed text and the preprocessed text in two separate space-split arrays. Make sure each non-alphabetical character also gets its own index in this array. Whatever list of words you're filtering, stem them also.
The next step would be to find the array indices which match to your list of stemmed 'stop' words. Remove those from the unprocessed array, and then rejoin on spaces.
This is only slightly more complicated, but will be much more reliable an approach. If you've got any doubts on the value of a more NLP-oriented approach, you might want to do some research into clbuttic mistakes.

can you try this:
https://regex101.com/r/P3qfKG/1
reg = /stop(\w+| [^ ]+|$)/gm
it will select both stop and start with stop and next word;

Javascript RegEx match 1-1-1 and 1-1-1-1-1 but not -1-1-1-1 or 1-1-1-1-

i haven't found anything when using google and stack overflow.
I need to match 1-1-1 but not -1-1-1 or 1-1-1- with javascript RegEx.
So it has to start with a number and end with a number and has to be seperated with "-".
I can't figure out, how to do it.
Is it even possible?

Unfortunately, JavaScript regex doesn't have a look-behind (see javascript regex - look behind alternative?), so to exclude a preceding -, the regex will have to match on the preceding character too (as long as it's not a -).
Since there might not be a preceding character (input starts with 1), you have to also match on beginning of input (^).
So, this regex will do it: (?:[^-]|^)(1(?:-1)+)(?!-)
See regex101.com.
Whether it should match a standalone 1, or only on 1-1 (and longer), is up to you. The regex above will not match standalone 1. Change + to * to change that.
I also added capturing of the actual text you wanted to match, i.e. without the leading character. You can remove the extra () around 1(?:-1)+ if that's not needed.

RegEx in JS to find No 3 Identical consecutive characters

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}

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.