I am having trouble making zero or one '?' give preference to one occurrence over zero, in javascript.
Ex. This is my regex: (=1)?
This is my string, str: abcd=1
when i do regex.exec(str) I get the following returned: ["",undefined]. It looks like it's choosing the zero length match in the beginning of my string. Is there a way to get it to choose '=1'? Possibly this may work differently in other languages but I'm currently using javascript, and this seems to be the case.
The expression (=1)? will give precedence to one occurrence over zero, but the regular expression engine will always attempt to match as early in the string as possible. So starting at the first character in the string, first it will try to match =1 and fail, and then because of the ? it will match the empty string.
I think the following expression is most similar to what your intention is:
(?:.*(=1))?
This will put =1 into the first capture group if it is anywhere in the string, but every string will still be matched because of the ? making the non-capturing group optional.
By defaut ? is greedy in javascript. Your problem is somewhere else.
aside note: to have ? lazy you must write ?? (like other quantifiers)
(=1)* will match =1=1=1=1=1=1=1=1
If you're looking to match all digits after the = then use
abcde=(\d+)
This will place all the digits into capture group 1.
Related
This is an example string:
123456#p654321
Currently, I am using this match to capture 123456 and 654321 in to two different groups:
([0-9].*)#p([0-9].*)
But on occasions, the #p654321 part of the string will not be there, so I will only want to capture the first group. I tried to make the second group "optional" by appending ? to it, which works, but only as long as there is a #p at the end of the remaining string.
What would be the best way to solve this problem?
You have the #p outside of the capturing group, which makes it a required piece of the result. You are also using the dot character (.) improperly. Dot (in most reg-ex variants) will match any character. Change it to:
([0-9]*)(?:#p([0-9]*))?
The (?:) syntax is how you get a non-capturing group. We then capture just the digits that you're interested in. Finally, we make the whole thing optional.
Also, most reg-ex variants have a \d character class for digits. So you could simplify even further:
(\d*)(?:#p(\d*))?
As another person has pointed out, the * operator could potentially match zero digits. To prevent this, use the + operator instead:
(\d+)(?:#p(\d+))?
Your regex will actually match no digits, because you've used * instead of +.
This is what (I think) you want:
(\d+)(?:#p(\d+))?
What does the regular expression, /(?!^)/g mean?
I can see that from here x(?!y): Matches x only if x is not followed by y.
That explains, if ?! is used before any set of characters, it checks for the not followed by condition.
But we have, ?!^. So, if we try to say it in words, it would probably mean not followed by negate. That really does not make me guess a probable statement for it. I mean negate of what?
I'm still guessing and could not reach a fruitful conclusion.
Help is much appreciated.
Thanks!
Circumflex ^ only implies negation in a character class [^...] (when comes as first character in class). Outside of it it means start of input string or line. A negative lookahead that contains ^ only, means match shouldn't be found at start of input string or line.
See it in action
It returns
all matches... (/g, global flag)
...which are not followed by... (x?!y, negative lookahead where x is the empty string)
...the position at the start of the string (^)
That is, any position between two characters except for the position at the start of the string, as you can see here.
This regex is useful, thus, for detecting empty strings (after all, applying the regex to an empty string will return no matches). Empty strings may be the result of splitting strings, and you probably don't want to do anything with them.
Usually in my system i have the following string:
http://localhost/api/module
to find out the last part of the string (which is my route) ive been using the following:
/[^\/]+$/g
However there may be cases where my string looks abit different such as:
http://localhost/api/module/123
Using the above regex it would then return 123. When my String looks like this i know that the last part will always be a number. So my question is how do i make sure that i can always find the last string that does not contain a number?
This is what i came up with which really stricty matches only module for the following lines:
http://localhost/api/module
http://localhost/api/module/123
http://localhost/api/module/123a
http://localhost/api/module/a123
http://localhost/api/module/a123a
http://localhost/api/module/1a3
(?!\w*\d\w*)[^\/][a-zA-Z]+(?=\/\w*\d+\w*|$)
Explanation
I basically just extended your expression with negative lookahead and lookbehind which basically matches your expression given both of the following conditions is true:
(?!\w*\d\w*) May contain letters, but no digits
[a-zA-Z]+ Really, truly only consists of one or more letters (was needed)
(?=\/\d+|$)The match is either followed by a slash, followed by digits or the end of the line
See this in action in my sample at Regex101.
partYouWant = urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
Here it is in action:
urlString="http://localhost/api/module/123"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
urlString="http://localhost/api/module"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
It just uses a capture expression to find the last non-numeric part.
It's going to do this too, not sure if this is what you want:
urlString="http://localhost/api/module/123/456"
urlString.replace(/^.*\/([a-zA-Z]+)[\/0-9]*$/,'$1')
-->"module"
/([0-9])\w+/g
That would select the numbers. You could use it remove that part from the url. What language are you using it for ?
I have to get the number of parenthesized substring matches in a regular expression:
var reg=/([A-Z]+?)(?:[a-z]*)(?:\([1-3]|[7-9]\))*([1-9]+)/g,
nbr=0;
//Some code
alert(nbr); //2
In the above example, the total is 2: only the first and the last couple of parentheses will create grouping matches.
How to know this number for any regular expressions?
My first idea was to check the value of RegExp.$1 to RegExp.$9, but even if there are no corresponding parenthseses, these values are not null, but empty string...
I've also seen the RegExp.lastMatch property, but this one represents only the value of the last matched characters, not the corresponding number.
So, I've tried to build another regular expression to scan any RegExp and count this number, but it's quite difficult...
Do you have a better solution to do that?
Thanks in advance!
Javascripts RegExp.match() method returns an Array of matches. You might just want to check the length of that result array.
var mystr = "Hello 42 world. This 11 is a string 105 with some 2 numbers 55";
var res = mystr.match(/\d+/g);
console.log( res.length );
Well, judging from the code snippet we can assume that the input pattern is always a valid regular expression, because otherwise it would fail before the some code partm right? That makes the task much easier!
Because We just need to count how many starting capturing parentheses there are!
var reg = /([A-Z]+?)(?:[a-z]*)(?:\([1-3]|[7-9]\))*([1-9]+)/g;
var nbr = (' '+reg.source).match(/[^\\](\\\\)*(?=\([^?])/g);
nbr = nbr ? nbr.length : 0;
alert(nbr); // 2
And here is a breakdown:
[^\\] Make sure we don't start the match with an escaping slash.
(\\\\)* And we can have any number of escaped slash before the starting parenthes.
(?= Look ahead. More on this later.
\( The starting parenthes we are looking for.
[^?] Make sure it is not followed by a question mark - which means it is capturing.
) End of look ahead
Why match with look ahead? To check that the parenthes is not an escaped entity, we need to capture what goes before it. No big deal here. We know JS doens't have look behind.
Problem is, if there are two starting parentheses sticking together, then once we capture the first parenthes the second parenthes would have nothing to back it up - its back has already been captured!
So to make sure a parenthes can be the starting base of the next one, we need to exclude it from the match.
And the space added to the source? It is there to be the back of the first character, in case it is a starting parenthes.
I need a form with one button and window for input
that will check an array, via a regular expression.
And will find a exact match of letters + numbers. Example wxyz [some space btw] 0960000
or a mix of numbers and letters [some space btw] + numbers 01xg [some space btw] 0960000
The array has four objects for now.
Once found i need a function the will open a new page or window when match is found .
Thanks you for your help.
Michael
To answer the Javascript part, here's one way to "grep" through the array to find matching elements:
var matches = [];
var re = /whatever/;
foo.forEach(
function(el) {
if( re.exec(el) )
matches.push(el);
}
);
To attempt to answer the regular expression part: I don't know what "exact match" means to you, and I'm assuming "some space" belongs only in between the other terms, and I'm assuming letters means the English alphabet from 'a' to 'z' in lower and upper case and the digits should be 0-9 (otherwise, other language characters might be matched).
The first pattern would be /[a-zA-Z0-9]+\s*0960000/. Change "\s*" to "\s+" if there is at least one space, instead of zero or more space characters. Change "\s" to " " if matching the tab character (and some lesser-used space chars) is not desirable.
For the second pattern, I don't know what "numbers 01xg" means, but if it means numbers followed by that string, then the pattern would be /[a-zA-Z0-9]+\s*[0-9]+\s*01xg\s*0960000/. The same caveats apply as above.
Additionally, this will match a partial string. If the string much be matched in entirety (if nothing in the string must exist except that which is matched), add "^" to the beginning of the pattern to anchor it to the beginning of the string, and "$" at the end to anchor it to the end of the string. For example, /[a-zA-Z0-9]+\s*0960000/ matches "foo_bar 5 0960000", but /^[a-zA-Z0-9]+\s*0960000$/ does not.
For more on regular expressions in Javascript, take a look at developer.mozilla.org's article on the RegExp object (the link takes you to JS version 1.5 reference, which should apply to all JS-capable browsers).
(edited to add): To match either situation, since they have overlapping parts, you could use the following pattern: /[a-zA-Z0-9]+(?:\s*[0-9]+\s*01xg)?\s*0960000/. The question mark says to match the part that differs -- in a non-matching group (?:foo) -- once or zero times. (?:foo)? and (?:foo|) do the same thing in this case, but I'm not sure whether there is a performance difference; I would recommend to use the one that makes the most sense to you, so you can read it later.