javascript regular expression for -999x999 - javascript

I need a regular expression for:
-[n digits]x[n digits]
I already tried this:
var s = "path/path/name-799x1024.jpg";
s.replace(/\d/g, "");
But this gets only the digits.
Here is a small jsfiddle: http://jsfiddle.net/aq6dp49n/
The outcome I try to get is:
pfad/pfade/name.jpg
How do I add the - and the small x between the two digits?

The regular expression that would match that is /-\d+x\d+/. Hence:
s.replace(/-\d+x\d+/, "")
Should work.
Here's what the regex means: the first - tells it that it should look for a - character. Then you have \d+ which means "one or more of \d", where \d is short-hand for the character class [0-9], i.e., all digits. After that you have x, which means it will look for the character x, and finally you have \d+ again, which is the same as before.

To match
-[n digits]x[n digits]
You would want
match(/-[0-9]{n}x[0-9]{n}\b/)
Though if you want an arbitrary (one or more) number of digits, use + in place of {n}. In the case of your example, you'd want 3 and 4 for your values of n.
Here's a step-by-step explanation of what this does:
/-[0-9]{3}x[0-9]{4}\b/
- matches the character - literally
[0-9]{3} match a single character present in the list below
Quantifier: {3} Exactly 3 times
0-9 a single character in the range between 0 and 9
x matches the character x literally (case sensitive)
[0-9]{4} match a single character present in the list below
Quantifier: {4} Exactly 4 times
0-9 a single character in the range between 0 and 9
\b assert position at a word boundary (^\w|\w$|\W\w|\w\W)
To remove the last size-like part of a string, this should do:
"path/path/name-799x1024.jpg".replace(/(.*)-[0-9]+x[0-9]+/, "$1");
// "path/path/name.jpg"
"path/path/name-10x12-799x1024.jpg".replace(/(.*)-[0-9]+x[0-9]+/, "$1");
// "path/path/name-10x12.jpg"
This takes advantage of the fact that regexps are greedy, so the (.*) absorbs (and saves) as much preceding text as possible before finding the next match.
(I prefer to use [0-9] in place of \d because it's more specific (\d also matches non-latin numerals) and therefore slightly faster, though in this case it shouldn't matter.)

Related

Javascript: Regex to exclude whitespace and special characters

I need a regex to validate,
Should be of length 18
First 5 characters should be either (xyz34|xyz12)
Remaining 13 characters should be alphanumeric only letters and numbers, no whitespace or special characters is allowed.
I have a pattern like here, '/^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13}/g'
But this is allowing whitespace and special characters like ($,% and etc) which is violating the rule #3.
Any suggestion to exclude this whitespace and special characters and to strictly check that it must be letters and numbers?
You should not quantify lookarounds. They are non-consuming patterns, i.e. the consecutive positive lookaheads check the presence of their patterns but do not advance the regex index, they check the text at the same position. It makes no sense repeating them 13 times. ^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13} is equal to ^(xyz34|xyz12)(?=.*[a-zA-Z])(?=.*[0-9]), and means the string can start with xyz34 or xyz12 and then should have at least 1 letter and at least 1 digits.
You may consider fixing the issue by using a consuming pattern like this:
If you do not care if the last 13 chars contain only digits or only letters, use the patterns suggested by other users, like /^(?:xyz34|xyz12)[a-zA-Z\d]{13}$/ or /^xyz(?:34|12)[a-zA-Z0-9]{13}$/
If there must be at least 1 digit and at least 1 letter among those 13 alphanumeric chars, use /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/.
See the regex demo #1 and the regex demo #2.
NOTE: these are regex literals, do not use them inside single- or double quotes!
Details
^ - start of string
xyz - a common prefix
(?:34|12) - a non-capturing group matching 34 or 12
(?=[a-zA-Z]*\d) - there must be at least 1 digit after any 0+ letters to the right of the current location
(?=\d*[a-zA-Z]) - there must be at least 1 letter after any 0+ digtis to the right of the current location
[a-zA-Z\d]{13} - 13 letters or digits
$ - end of string.
JS demo:
var strs = ['xyz34abcdefghijkl1','xyz341bcdefghijklm','xyz34abcdefghijklm','xyz341234567890123','xyz14a234567890123'];
var rx = /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/;
for (var s of strs) {
console.log(s, "=>", rx.test(s));
}
.* will match any string, for your requirment you can use this:
/^xyz(34|12)[a-zA-Z0-9]{13}$/g
regex fiddle
/^(xyz34|xyz12)[a-zA-Z0-9]{13}$/
This should work,
^ asserts position at the start of a line
1st Capturing Group (xyz34|xyz12)
1st Alternative xyz34 matches the characters xyz34 literally (case sensitive)
2nd Alternative xyz12 matches the characters xyz12 literally (case sensitive)
Match a single character present in the list below [a-zA-Z0-9]{13}
{13} Quantifier — Matches exactly 13 times

How does the following code mean two consecutive numbers?

This is from an exercise on FCC beta and i can not understand how the following code means two consecutive numbers seeing how \D* means NOT 0 or more numbers and \d means number, so how does this accumulate to two numbers in a regexp?
let checkPass = /(?=\w{5,})(?=\D*\d)/;
This does not match two numbers. It doesn't really match anything except an empty string, as there is nothing preceding the lookup.
If you want to match two digits, you can do something like this:
(\d)(\d)
Or if you really want to do a positive lookup with the (?=\D*\d) section, you will have to do something like this:
\d(?=\D*\d)
This will match against the last digit which is followed by a bunch of non-digits and a single digit. A few examples (matched numbers highlighted):
2 hhebuehi3
^
245673
^^^^^
2v jugn45
^ ^
To also capture the second digit, you will have to put brackets around both numbers. Ie:
(\d)(?=\D*(\d))
Here it is in action.
In order to do what your original example wants, ie:
number
5+ \w characters
a non-number character
a number
... you will need to precede your original example with a \d character. This means that your lookups will actually match something which isn't just an empty string:
\d(?=\w{5,})(?=\D*\d)
IMPORTANT EDIT
After playing around a bit more with a JavaScript online console, I have worked out the problem with your original Regex.
This matches a string with 5 or more characters, including at least 1 number. This can match two numbers, but it can also match 1 number, 3 numbers, 12 numbers, etc. In order to match exactly two numbers in a string of 5-or-more characters, you should specify the number of digits you want in the second half of your lookup:
let regex = /(?=\w{5,})(?=\D*\d{2})/;
let string1 = "abcd2";
let regex1 = /(?=\w{5,})(?=\D*\d)/;
console.log("string 1 & regex 1: " + regex1.test(string1));
let regex2 = /(?=\w{5,})(?=\D*\d{2})/;
console.log("string 1 & regex 2: " + regex2.test(string1));
let string2 = "abcd23";
console.log("string 2 & regex 2: " + regex2.test(string2));
My original answer was about Regex in a vacuum and I glossed over the fact that you were using Regex in conjunction with JavaScript, which works a little differently when comparing Regex to a string. I still don't know why your original answer was supposed to match two numbers, but I hope this is a bit more helpful.
?= Positive lookahead
w{5,} matches any word character (equal to [a-zA-Z0-9_])
{5,}. matches between 5 and unlimited
\D* matches any character that\'s not a digit (equal to [^0-9])
* matches between zero and unlimited
\d matches a digit (equal to [0-9])
This expression is global - so tries to match all
You can always check your expression using regex101

Regex allowing no leading whitespace but allowing anywhere else and requiring at least one uppercase and one lowercase letters

I want regex code to validate usernames that have:
length between 6 and 30
contain at least one letter from A-Z
contain at least one digit from 0-9
not contain a space at the beginning but it might have at
the end or in the middle.
may contain special characters
So far I have tried this:
^[\S](?=.*\d)(?=.*[A-Z]).{6,30}$
It works quite good but when I choose an uppercase letter ONLY at the beginning it doesnt validate my password.
Test12 34 ----> Doesnt accept but should accept
TesT12 34 ----> Accept
tesT12 34 ----> Accept
The problem arises because the \S is at the pattern start before the lookaheads. That means, that the lookaheads that require an uppercase ASCII letter and a digit only check them after the first character in string.
Now, there can be two scenarios: 1) there may be any amount of whitespace characters in the string, but at its beginning, 2) only one whitespace char is allowed in the string, and not in the beginning
Scenario 1
Put \S after the lookaheads and decrement the limiting quantifier values to set the limits to 6-30 as \S already matches and consumes the first char:
^(?=.*\d)(?=.*[A-Z])\S.{5,29}$
^^ ^^^^
See the regex demo.
JS test:
var rx = /^(?=.*\d)(?=.*[A-Z])\S.{5,29}$/;
var vals = [ "Test12 34", "TesT12 34", "tesT12 34" ];
for (var s of vals) {
console.log(s, "=>", rx.test(s));
}
Scenario 2
Use
^(?=.{6,30}$)(?=.*\d)(?=.*[A-Z])\S+(?:\s\S*)?$
The length is restricted with the positive lookahead at the beginning ((?=.{6,30}$)) and the consuming \S+(?:\s\S*)? pattern will only allow a single \s whitespace (1 or 0 due to the last ? - one or zero occurrences quantifier) and it can be in the middle (as the first \S is quantified with +, one or more occurrences quantifier) or end (as the \S after \s is *-quantified, zero or more occurrences quantifier).
See the regex demo.

JavaScript regular expressions to match no digits, whitespace and selected symbols

Thanks for taking a look.
My goal is to come up with a regexp that will match input that contains no digits, whitespace or the symbols !#£$%^&*()+= or any other symbol I may choose.
I am however struggling to grasp precisely how regular expressions work.
I started out with the simple pattern /\D/, which from my understanding will match the first non-digit character it can find. This would match the string 'James' which is correct but also 'James1' which I don't want.
So, my understanding is that if I want to ensure that a pattern is not found anywhere in a given string, I use the ^ and $ characters, as in /^\D$/. Now because this will only match a single character that is not a digit, I needed to use + to specify that 1 or more digits should not be founds in the entire string, giving me the expression /^\D+$/. Brilliant, it no longer matches 'James1'.
Question 1
Is my reasoning up to this point correct?
The next requirement was to ensure no whitespace is in the given string. \s will match a single whitespace and [^\s] will match the first non-whitespace character. So, from my understanding I just had to add this to what I have already to match strings that contain no digits and no whitespace. Again, because [^\s] will only match a single non-white space character, I used + to match one or more whitespace characters, giving the new regexp of /^\D+[^\s]+$/.
This is where I got lost, as the expression now matches 'James1' or even 'James Smith25'. What? Massively confused at this point.
Question 2
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Question 3
How would I go about writing the regular expression I'm trying to solve?
While I am keen to solve the problem I am more interested in figuring where my understanding of regular expressions is lacking, so any explanations would be helpful.
Not quite; ^ and $ are actually "anchors" - they mean "start" and "end", it's actually a little more complicated, but you can consider them to mean the start and end of a line for now - look up the various modifiers on regular expressions if you're interested in learning more about this. Unfortunately ^ has an overloaded meaning; if used inside square brackets it means "not", which is the meaning you are already acquainted with. It's very important that you understand the difference between these two meanings and that the definition in your head actually applies only to character range matching!
Contributing further to your confusion is that \d means "a numerical digit" and \D means "not a numerical digit". Similarly \s means "a whitespace (space/tab/newline/etc.) character" and \S means "not a whitespace character."
It's worth noting that \d is effectively a shortcut for [0-9] (note that - has a special meaning inside square brackets), and \D is a shortcut for [^0-9].
The reason it's matching strings that contain spaces is that you've asked for "1+ non-numerical digits followed by 1+ non-space characters" - so it'll match lots of strings! I think that perhaps you don't understand that regular expressions match bits of strings, you're not adding constraints as you go, but rather building up bots of matchers that will match bits of corresponding strings.
/^[^\d\s!#£$%^&*()+=]+$/ is the answer you're looking for - I'd look at it like this:
i. [] - match a range of characters
ii. []+ - match one or more of that range of characters
iii. [^\d\s]+ - match one or more characters that do not match \d (numerical digit) or \s (whitespace)
iv. [^\d\s!#£$%^&*()+=]+ - here's a bunch of other characters I don't want you to match
v. ^[^\d\s!#£$%^&*()+=]+$ - now there are anchors applied, so this matcher has to apply to the whole line otherwise it fails to match
A useful website to explore regexs is http://regexr.com/3b9h7 - which I supply with my suggested solution as an example. Edit: Pruthvi Raj's link to debuggerx is awesome!
Is my reasoning up to this point correct?
Almost. /\D/ matches any character other than a digit, but not just the first one (if you use g option).
and [^\s] will match the first non-whitespace character
Almost, [^\s] will match any non-whitespace character, not just the first one (if you use g option).
/^\D+[^\s]+$/ matching strings that contain spaces?
Yes, it does, because \D matches a space (space is not a digit).
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Because \D+ in /^\D+[^\s]+$/can match spaces.
Conclusion:
Use
^[^\d\s!#£$%^&*()+=]+$
It will match strings that have no digits and spaces, and the symbols you do not allow.
Mind that to match a literal -, ] or [ with a character class, you either need to escape them, or use at the start or end of the expression. To play it safe, escape them.
Just insert every character you don't want to include in a negated character class as follows:
^[^\s\d!#£$%^&*()+=]*$
DEMO
Debuggex Demo
^ - start of the string
[^...] - matches one character that is not in `...`
\s - matches a whitespace (space, newline,tab)
\d - matches a digit from 0 to 9
* - a quantifier that repeats immediately preceeding element by 0 or more times
so the regex matches any string that has
1. string that has a beginning
2. containing 0 or more number of characters that is not whitesapce, digit, and all the symbols included in the character class ( In this example !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class

Regex improvement match

I need some help to improve a regex!
In JavaScript I have a regular expression which looks for pairs of numbers in a filename
var nums = str.match(/[\d]{1,}[\d]{1,}/gi);
This will match
DV_Banner_1200x627.jpg
DV_Banner_1200y627.jpg
DV_Banner_1200 x 627.jpg
DV_Banner_1200 x627.jpg
DV_Banner_1200 627.jpg
with (1200,627)
I have tried to improve the reg ex, just incase there are more than two pairs of numbers, to look for the following
number(1 digit or more) + whitspace(1 or more) + x (zero or once) + whitspace(1 or more) + number(1 digit or more)
Which should fail on the second example (using a 'y' instead on an 'x'), which I thought would be:
[\d]{1,}[\s]?[x]?[\s]?[\d]{1,}
but it grabs all the digits in
DV_Banner_1200 x 627 01.jpg
with (1200,627,01) whereas I only want the first two numbers. I've written the code to deal only with the first two, but I was wondering where I was going wrong. Only a level 17 regex wizard can save me now! Thanks
I used \d+\s?x?\s?\d+ as my regex (same thing just replacing + for {1,} and removing the unnecessary []). You can see the outcome of it here.
The reason it's matching the 01 is because of all the ?. So it's matching the first /d+ (1 digit: 0), and then 0 of \s, 0 of x, and 0 of \s followed by \d+ (another 1 digit: 1)
The regex
(\d+)(?:\s?x\s?|\s)(\d+)
should do the trick. Test it here
(?:...) is a non-capture group. So it allows alternation while not assigning a back reference to it. This part matches the characters in between the two numbers (either has an x or a <space>).
Just try with following regex:
(\d+)(?:(?: ?x ?)| )(\d+)
demo
You say you want "one or more" whitespace characters between the "x", but you have used the ? quantifier which means "zero or one". Thus, because you've also marked the "x" as optional, it will match any two-or-more digit number: Your first [\d]{1,} will match against 0 then your second one will match on 1.
Note that you do not need to enclose single atoms into a character range: [\d] can be more simply written as \d. Also {1,} -- meaning "one or more" -- is more easily encoded as +.
As you want "one or more" whitespace character on either side of the "x", I would go with:
\d+(?:(?:\s+x\s+)|\s+)\d+
Note that (?: ... ) is a "non-capture group", so these bits won't form part of your match array. However, I don't think you want "one or more" whitespace character, as that won't match your first example. Instead, try this:
\d+(?:(?:\s*x\s*)|\s+)\d+
Where the * quantifier means "zero-or-more".

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