Like the title says, I have a (faulty) Regex in JavaScript, that should check for a "2" character (in this case) surrounded by slashes. So if the URL was http://localhost/page/2/ the Regex would pass.
In my case I have something like http://localhost/?page=2 and the Regex still passes.
I'm not sure why. Could anyone tell me what's wrong with it?
/^(.*?)\b2\b(.*?$)/
(I'm going to tell you, I didn't write this code and I have no idea how it works, cause I'm really bad with Regex)
Seems too simple but shouldn't this work?:
/\/2\//
http://jsfiddle.net/QHac8/1/
As it's javascript you have to escape the forward slashes as they are the delimiters for a regex string.
or if you want to match any number:
/\/\d+\//
You don't check for a digit surrounded by slashes. The slashes you see are only your regex delimiters. You check for a 2 with a word boundary \b on each side. This is true for /2/ but also for =2
If you want to allow only a 2 surrounded by slashes try this
/^(.*?)\/2\/(.*?)$/
^ means match from the start of the string
$ match till the end of the string
(.*?) those parts are matching everything before and after your 2 and those parts are stored in capturing groups.
If you don't need those parts, then Richard D is right and the regex /\/2\// is fine for you.
Related
I have 3 URL scenarios
normal urls:
http://example.com,
https://example.com,
https://example.com/products/1234,
https://example.com/products?category=shoes&limit=10
url with context variable:
https://example.com/products/{$context.productId}
whole url is context variable:
{$varContext.getProductUrl}
I have to validate input URL to match above 3 scenarios.
I have written regex but which is not working for all scenarios.
^((?:http(s){0,1}://[\\w\\-\\.:/\$\{\}=\?&]+)|(?:\{\$[a-zA-Z]+\.[a-zA-Z]+\}))$
Could any one please help with this?
Hi you made a very simple mistake. Stopped 1% away from the finish line.
^((?:http(s){0,1}:\/\/[\w\-\.:/\${}=\?&]+)|(?:{\$[a-zA-Z]+.[a-zA-Z]+}))$
is the correct one you just forgot to escape the / to denote (mean) a literal /. Mistake appears right after ^((?:http(s){0,1}:
Next time use a site like Regex101, a site to help you test your regex in case you didn't know.
Apart from escaping the forward slashes after http://, you are escaping the backslash in the character class. \\w would match a backslash and a w. This part \\-\\ would match a range from \ till \ which would match a backslash.
That would match for example http://w\.:/${}=?&
As an addition, you could make a few adjustments to your regex.
(s){0,1} can be written as https?
The character class [\\w\\-\\.:/\$\{\}=\?&] can be written as [\w.:/${}=?&-] without the escaping of .$? and the hyphen can be moved to the beginning or at the end. \\ would match a backslash.
If you don't need the capturing group () you could omit that and only use an alternation |
You could use the /i to get a case insensitive match and the character class would become [a-z]
Escape the dot \. to match it literally
For example:
const regex = /^https?:\/\/[\w.:/${}=?&-]+|{\$[a-z]+\.[a-z]+}$/i;
See the Regex demo
I'm been having trouble with regex, which I doesn't understand at all.
I have a string '#anything#that#i#say' and want that the regex detect one word per #, so it will be [#anything, #that, #i, #say].
Need to work with spaces too :(
The closest that I came is [#\w]+, but this only get 1 word and I want separated.
You're close; [#\w] will match anything that is either a # or a word character. But what you want is to match a single # followed by any number of word characters, like this: #\w+ without the brackets
var str = "#anything#that#i#say";
var regexp = /#\w+/gi;
console.log(str.match(regexp));
It's possible to have this deal with spaces as well, but I'd need to see an example of what you mean to tell you how; there are lots of ways that "need to work with spaces" can be interpreted, and I'd rather not guess.
use expression >> /#\s*(\w+)/g
\s* : to check if zero or more spaces you have between # and word
This will match 4 word in your string '#anything#that#i#say'
even your string is containing space between '#anything# that#i# say'
sample to check: http://www.regextester.com/?fam=97638
For example /(www\.)?(.+)(\.com)?/.exec("www.something.com") will result with 'something.com' at index 1 of the resulting array. But what if we want to capture only 'something' in a capturing group?
Clarifications:
The above string is just for example - we dont want to assume anything about the suffix string (.com above). It could as well be orange.
Just this part can be solved in C# by matching from right to left (I dont know of a way of doing that in JS though) but that will end up having www. included then!
Sure, this problem as such is easily solvable mixing regex with other string methods like replace / substring. But is there a solution with only regex?
(?:www\.)?(.+?)(?:\.com|$)
This will give only something ingroups.Just make other groups non capturing.See demo.
https://regex101.com/r/rO0yD8/4
Just removing the last character (?) from the regex does the trick:
https://regex101.com/r/uR0iD2/1
The last ? allows a valid output without the (\.com) matching anything, so the (.+) can match all the characters after the www..
Another option is to replace the greedy quantifier +, which always tries to match as much characters as possible, with the +?, which tries to match as less characters as possible:
(www\.)?(.+?)(\.com)?$
https://regex101.com/r/oY7fE0/2
Note that it is necessary to force a match with the entire string through the end of line anchor ($).
If you only want to capture "something", use non-capturing groups for the other sections:
/(?:www\.)?(.+)(?:\.com)?/.exec("www.something.com")
The ?: denotes the groups as non-capturing.
When I use a tool like regexpal.com it let's me use regex as I am used to. So for example I want to check a text if there is a match for a word that is at least 3 letters long and ends with a white space so it will match 'now ', 'noww ' and so on.
On regexpal.com this regex works \w{3,}\s this matches both the words above.
But on javascript I have to add double backslashes before w and s. Like this:
var regexp = new RegExp('\\w{3,}\\s','i');
or else it does not work. I looked around for answers and searched for double backslash javascript regex but all I got was completely different topics about how to escape backslash and so on. Does someone have an explanation for this?
You could write the regex without double backslash but you need to put the regex inside forward slashshes as delimiter.
/^\w{3,}\s$/.test('foo ')
Anchors ^ (matches the start of the line boundary), $ (matches the end of a line) helps to do an exact string match. You don't need an i modifier since \w matches both upper and lower case letters.
Why? Because in a string, "\" quotes the following character so "\w" is seen as "w". It essentially says "treat the next character literally and don't interpret it".
To avoid that, the "\" must be quoted too, so "\\w" is seen by the regular expression parser as "\w".
I just wanted to ask for an example of a string that would match this regular expression for JS:
/\/[a-z]{2,6}\/(\([0-9]+\)?$/
But this bit confuses me: /(\([0-9]+\)?$/
If I could get an example of a string using this regular expression and a brief explanation, that would be enough to clear it up for me.
Thanks!
EDITED: Sorry for the trouble, I missed a parentheses, however I just want to clear up, a string that would match would be such as:
/ab/(12345) or /abcdef/(1) etc right?
Here is a visual representation of the #Paul Roub explanation:
EDITED with last pattern
\/[a-z]{2,6}\/\([0-9]+\)$
Debuggex Demo
This example is OK : /abcdef/(12345)
\/
A forward slash (escaped), followed by
[a-z]{2,6}
between two and six lowercase English letters, followed by
\/
another slash
([0-9]+)?
the inside part - one or more digits. ? means "zero or one", so we're looking for a string of digits, or nothing. The parentheses would let this number be captured as a group for later processing
$
and the end of the string.
Things that would match:
/ab/0
/ab/
/acdefg/12345
things that would not match
/a/0
/abcdefgh/12345
/ab/0x
That regexp is malformed.
Opening bracket is escaped, but closing one not
Opening paren is not escaped but closing one is
I think the intended regexp was:
/\/[a-z]{2,6}\/([0-9]+)?$/
This would match:
/ab/1
/ab/
/abc/123
Wouldn't match:
/a/1
ab/1
/abcdefg/123
Cheers.