[\w+\.]{3}
and
\w+\.\w+\.\w+\.
the former matches "dra"
later matches "dragon.is.awesome"
What am I not understanding right about them?
Input text looks like
i know dragon.is.awesome but
i know dragon.is.awesome.because, he is awesome
i know dragon.sucks.because, he is not awesome
i know dragon.is.dead, someone killed him
so i need to match any combination of groupings that are of the pattern \w+.
Because the first one is a character class.
[\w+/\.]
matches either one \w, or one + or one / or one literal .. If you want to shorten the latter, use normal parentheses:
(\w+\.){3}
Note that within character classes, most meta-characters lose their meaning. So + and . and * (for example) can all be contained and matched without being escaped.
[...] is a character class. It matches one character. [\w+\.] matches one character which is either a "word" character (letter, number, or underscore), or a plus, or a dot. [\w+\.]{3} matches three such characters in a row.
[] is a character class, not a subpattern. [abc] Matches a single a, b or c.
You probably meant (\w+\.){3}, which does match the same as your second regex.
Related
I'm trying to create a regex that checks if a string is a valid path for Firestore document.
I will find a regex that testing if a string:
start with a char ^([a-z]{1})
after first char, there will be only letter/digit and/or a dot \w*(.?\w+){0,}
last chars in the string could be an index of an array (\[{1}\d+\]{1})?$
First and second points work well but the last group doesn't work. I test a string like data.images[11 and the regex return true.
first of all you can shorten some quantifiers in your regex:
{1} -> can be ignored completely
{0,} -> *
Your second part could be expressed like this, this will also support readability:
[\w.]* meaning: take any character inside the brackets 0 to n-times. The bracket expression also supports predefined classes, so we are using \w here. The dot INSIDE the brackets doesn't need to be escaped, it simply means the one character dot.
So your parts would be:
^([a-z])
[\w.]*
(\[\d+\])?$
I hope this helps. According to regexpal it matches data.images[11], but not data.images[11. Also it seems to support all your demands.
EDIT:
Your second part doesn't work because (like Asocia stated in the answer) you would need to escape the dot. The dot itself is a class meaning "any character" (depending on regex engine and settings sometimes even line breaks). As you mean the dot as a character you need to escape it.
Yet another question about a regex.
I'm trying to match all special characters, except '*'.
So if I match my regex against:
John%%%* dadidou
I should get:
John* dadidou
Here: How to match with regex all special chars except "-" in PHP?
The accepted answer advices to use (if I want to exclude '-'):
[^\w-]
But doesn't that mean: "NOT a special character, NOT -", which is a bit redundant ?
What you really want is this regex for matching:
[^\w\s*]+
Replace it by empty string.
Which means match 1 or more of any character that is:
Not a word character [AND]
Not a whitespace [AND]
Not a literal *
RegEx Demo
When you define a negative character class, you are really inverting it.
What does that mean ?
A positive character class implicitly OR's it's contents.
When you negate a class, you implicitly AND it's contents.
So, [\w-] means word OR dash,
the inverse, [^\w-] means not word AND not dash.
A negative word for instance, [^\w] would match a dash -.
So, to not match it, you have to add a not dash as well.
A C analogy would be
existing (varA || varB)
inverted (!varA && !varB)
where inverting changes the Boolean of each of the components.
Basically a negative class changes the Boolean of each of its components,
so the implicit OR becomes an implicit AND and the components characters
(or expressions) are negated.
What will really bake your noodle later on is when you see something like
[^\S\r\n]
This translates to NOT-NOT-Whitespace and NOT-cr and NOT-lf
which reduces to matching all whitespace except CR,LF
I am looking to clean up a regular expression which matches 2 or more characters at a time in a sequence. I have made one which works, but I was looking for something shorter, if possible.
Currently, it looks like this for every character that I want to search for:
([A]{2,}|[B]{2,}|[C]{2,}|[D]{2,}|[E]{2,}|...)*
Example input:
AABBBBBBCCCCAAAAAADD
See this question, which I think was asking the same thing you are asking. You want to write a regex that will match 2 or more of the same character. Let's say the characters you are looking for are just capital letters, [A-Z]. You can do this by matching one character in that set and grouping it by putting it in parentheses, then matching that group using the reference \1 and saying you want two or more of that "group" (which is really just the one character that it matched).
([A-Z])\1{1,}
The reason it's {1,} and not {2,} is that the first character was already matched by the set [A-Z].
Not sure I understand your needs but, how about:
[A-E]{2,}
This is the same as yours but shorter.
But if you want multiple occurrences of each letter:
(?:([A-Z])\1+)+
where ([A-Z]) matches one capital letter and store it in group 1
\1 is a backreference that repeats group 1
+ assume that are one or more repetition
Finally it matches strings like the one you've given: AABBBBBBCCCCAAAAAADD
To be sure there're no other characters in the string, you have to anchor the regex:
^(?:([A-Z])\1+)+$
And, if you wnat to match case insensitive:
^(?i)(?:([A-Z])\1+)+$
I know it seems a bit redundant but I'd like a regex to match anything.
At the moment we are using ^*$ but it doesn't seem to match no matter what the text.
I do a manual check for no text but the test view we use is always validated with a regex. However, sometimes we need it to validate anything using a regex. i.e. it doesn't matter what is in the text field, it can be anything.
I don't actually produce the regex and I'm a complete beginner with them.
The regex .* will match anything (including the empty string, as Junuxx points out).
The chosen answer is slightly incorrect, as it wont match line breaks or returns. This regex to match anything is useful if your desired selection includes any line breaks:
[\s\S]+
[\s\S] matches a character that is either a whitespace character (including line break characters), or a character that is not a whitespace character. Since all characters are either whitespace or non-whitespace, this character class matches any character. the + matches one or more of the preceding expression
^ is the beginning-of-line anchor, so it will be a "zero-width match," meaning it won't match any actual characters (and the first character matched after the ^ will be the first character of the string). Similarly, $ is the end-of-line anchor.
* is a quantifier. It will not by itself match anything; it only indicates how many times a portion of the pattern can be matched. Specifically, it indicates that the previous "atom" (that is, the previous character or the previous parenthesized sub-pattern) can match any number of times.
To actually match some set of characters, you need to use a character class. As RichieHindle pointed out, the character class you need here is ., which represents any character except newlines (and it can be made to match newlines as well using the appropriate flag). So .* represents * (any number) matches on . (any character). Similarly, .+ represents + (at least one) matches on . (any character).
I know this is a bit old post, but we can have different ways like :
.*
(.*?)
I'd like a regular expression to match a string only if it contains a character that occurs a predefined number of times.
For example:
I want to match all strings that contain the character "_" 3 times;
So
"a_b_c_d" would pass
"a_b" would fail
"a_b_c_d_e" would fail
Does someone know a simple regular expression that would satisfy this?
Thank you
For your example, you could do:
\b[a-z]*(_[a-z]*){3}[a-z]*\b
(with an ignore case flag).
You can play with it here
It says "match 0 or more letters, followed by '_[a-z]*' exactly three times, followed by 0 or more letters". The \b means "word boundary", ie "match a whole word".
Since I've used '*' this will match if there are exactly three "_" in the word regardless of whether it appears at the start or end of the word - you can modify it otherwise.
Also, I've assumed you want to match all words in a string with exactly three "_" in it.
That means the string "a_b a_b_c_d" would say that "a_b_c_d" passed (but "a_b" fails).
If you mean that globally across the entire string you only want three "_" to appear, then use:
^[^_]*(_[^_]*){3}[^_]*$
This anchors the regex at the start of the string and goes to the end, making sure there are only three occurences of "_" in it.
Elaborating on Rado's answer, which is so far the most polyvalent but could be a pain to write if there are more occurrences to match :
^([^_]*_){3}[^_]*$
It will match entire strings (from the beginning ^ to the end $) in which there are exactly 3 ({3}) times the pattern consisting of 0 or more (*) times any character not being underscore ([^_]) and one underscore (_), the whole being followed by 0 ore more times any character other than underscore ([^_]*, again).
Of course one could alternatively group the other way round, as in our case the pattern is symmetric :
^[^_]*(_[^_]*){3}$
This should do it:
^[^_]*_[^_]*_[^_]*_[^_]*$
If you're examples are the only possibilities (like a_b_c_...), then the others are fine, but I wrote one that will handle some other possibilities. Such as:
a__b_adf
a_b_asfdasdfasfdasdfasf_asdfasfd
___
_a_b_b
Etc.
Here's my regex.
\b(_[^_]*|[^_]*_|_){3}\b