Develop Reference

JS Regex lookahead

I want to be able to match these types of strings (comma separated, and no beginning or trailing spaces):
LaunchTool[0].Label,LaunchTool[0].URI,LaunchTool[1].Label,LaunchTool[1].URI,LaunchItg[0].Label,LaunchItg[0].URI,csr_description
The rules, in English, are:
1) Zero or more instances of [] where the brackets must contain only one number 0-9
2) Zero or more instances of ., where . must be followed by a letter
3) Zero or more instances of _, where _ must be followed by a letter
I currently have this regex:
/^([a-z]){1,}(\[[0-9]\]){0,}(\.){0,}[a-z]{1,}$/i
I cannot figure out why
"aaaa" doesn't match
furthemore,
"aaaa[0].a" matches, but "aaaa[0]" does not...
anyone know what's wrong? I believe I might need a lookahead to make sure . and _ characters are followed by a letter? Perhaps I can avoid it.

this regex can match "aaaa", try getting value of
(/^([a-z]){1,}(\[[0-9]\]){0,}(\.){0,}[a-z]{1,}$/i).test("aaaa")
"aaaa[0]" does not match, because there is [a-z]{1,} in the end of expression. once "[0]" is matched by (\[[0-9]\]){0,}, trailing [a-z]{1,} must be shown at the end of string

Use optional capture groups. Example: ([a-z])?.
Here is what i ended up with:
/^((\w+)(\[\d+\])?\.?(\w+)?,?)+$/
Shorthands explanation:
* = {0,}
+ = {1,}
\w = [A-Za-z0-9_]
\d = [0-9]

Regex - match variable beginned with $ in javascript?

I want to match all the variables like $bar, $foo, I only have this until the moment:
(\$)+[A-Za-z]

Just this:
/\$[\w_]+/g
And if you want to check the variable does not start with number, just check by negative lookahead. (?![\d])
/\$(?![\d])[\w_]+/g
/g to match all.

This should match some common valid variables which consists of [A-Za-z0-9_] but don't start with number.
/(\$(?!\d)\w+)/g
/ .. /g // regular expression is put between two //, the g behind is flag for global
( .. ) // a capturing group, can be called using \1 or $1 depending on regex processor
\$ // escaped character `$`
(?!\d) // negative lookahead - ensures that next character after $ isn't a match ->
-> \d // matches one digit
\w+ // one or more "word characters". Matches the ASCII characters [A-Za-z0-9_]
note: if you are going to match all variable names globally, you don't need any capturing groups, so just use /\$(?!\d)\w+/g.
According to php.net, this is the right regexp: \$+[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*
EDIT: I have just noticed you are not matching PHP variables at all, but you get the idea now. :)
For fun http://regex.alf.nu/

\$[A-Za-z]+\w*
Explained: $ in the start, one or more letters, followed by 0 or more valid variable letters (depending on language).
See https://www.regex101.com/r/cB1bM1/1 for a detailed explanation.

The regex you may need is:
\$\w+

Regex detect chars duplicates

I use expression to detect duplicate one char in string.
if (str.replace(/[^#]/g, '').length > 1) {...}
Now, i wish detect duplicate more than one char.
For sample, lets try detect duplicates of # or #:
Wordwithone#here - match
Word#withduplicate#here - not
Word#withoneandone#here - match
Is it possible with regexp detect duplicate char from set of chars? (## as in sample)

Your attempt in the comment /([#])\1{2,}/ig will search for consecutive # in the string.
It only needs a simple modification to make it work:
/([##])[\s\S]*?\1/.test(inputString)
Either [\s\S]*? or [\s\S]* would work, just that the order of searching would be a bit different:
*? is lazy, so the search for duplicate will start from the adjacent character.
* is greedy, so the search will start from the last character in the string.
The point is to allow any number of characters to come in-between the 2 duplicates [\s\S]*. I used [\s\S] to allow any character, as opposed to . which excludes the following new line characters: \n, \r, \u2028 and \u2029.

Using regular expression in Javascript

I need to check whether information entered are 3 character long, first one should be 0-9 second A-Z and third 0-9 again.
I have written pattern as below:
var pattern = `'^[A-Z]+[0-9]+[A-Z]$'`;
var valid = str.match(pattern);
I got confused with usage of regex for selecting, matching and replacing.
In this case, does[A-Z] check only one character or whole string ?
Does + separate(split?) out characters?

1) + matches one or more. You want exactly one
2) declare your pattern as a REGEX literal, inside forward slashes
With these two points in mind, your pattern should be
/^[A-Z][0-9][A-Z]$/
Note also you can make the pattern slightly shorter by replacing [0-9] with the \d shortcut (matches any numerical character).
3) Optionally, add the case-insensitive i flag after the final trailing slash if you want to allow either case.
4) If you want to merely test a string matches a pattern, rather than retrieve a match from it, use test(), not match() - it's more efficient.
var valid = pattern.test(str); //true or false

+ means one or more characters so a possible String would be ABCD1234EF or A3B, invalid is 3B or A 6B

This is the regex you need :
^[0-9][A-Z][0-9]$
In this case, does[A-Z] check only one character or whole string ?
It's just check 1 char but a char can be many times in a string..
you should add ^ and $ in order to match the whole string like I did.
Does + separate(split?) out characters?
no.
+ sign just shows that a chars can repeat 1+ times.

"+" means one or more. In your case you should use exact quantity match:
/^\w{1}\d{1}\w{1}$/

Javascript multiple regex pattern

I'm trying to exclude some internal IP addresses and some internal IP address formats from viewing certain logos and links in the site.I have multiple range of IP addresses(sample given below). Is it possible to write a regex that could match all the IP addresses in the list below using javascript?
10.X.X.X
12.122.X.X
12.211.X.X
64.X.X.X
64.23.X.X
74.23.211.92
and 10 more

Quote the periods, replace the X's with \d+, and join them all together with pipes:
const allowedIPpatterns = [
"10.X.X.X",
"12.122.X.X",
"12.211.X.X",
"64.X.X.X",
"64.23.X.X",
"74.23.211.92" //, etc.
];
const allowedRegexStr = '^(?:' +
allowedIPpatterns.
join('|').
replace(/\./g, '\\.').
replace(/X/g, '\\d+') +
')$';
const allowedRegexp = new RegExp(allowedRegexStr);
Then you're all set:
'10.1.2.3'.match(allowedRegexp) // => ['10.1.2.3']
'100.1.2.3'.match(allowedRegexp) // => null
How it works:
First, we have to turn the individual IP patterns into regular expressions matching their intent. One regular expression for "all IPs of the form '12.122.X.X'" is this:
^12\.122\.\d+\.\d+$
^ means the match has to start at the beginning of the string; otherwise, 112.122.X.X IPs would also match.
12 etc: digits match themselves
\.: a period in a regex matches any character at all; we want literal periods, so we put a backslash in front.
\d: shorthand for [0-9]; matches any digit.
+: means "1 or more" - 1 or more digits, in this case.
$: similarly to ^, this means the match has to end at the end of the string.
So, we turn the IP patterns into regexes like that. For an individual pattern you could use code like this:
const regexStr = `^` + ipXpattern.
replace(/\./g, '\\.').
replace(/X/g, '\\d+') +
`$`;
Which just replaces all .s with \. and Xs with \d+ and sticks the ^ and $ on the ends.
(Note the doubled backslashes; both string parsing and regex parsing use backslashes, so wherever we want a literal one to make it past the string parser to the regular expression parser, we have to double it.)
In a regular expression, the alternation this|that matches anything that matches either this or that. So we can check for a match against all the IP's at once if we to turn the list into a single regex of the form re1|re2|re3|...|relast.
Then we can do some refactoring to make the regex matcher's job easier; in this case, since all the regexes are going to have ^...$, we can move those constraints out of the individual regexes and put them on the whole thing: ^(10\.\d+\.\d+\.\d+|12\.122\.\d+\.\d+|...)$. The parentheses keep the ^ from being only part of the first pattern and $ from being only part of the last. But since plain parentheses capture as well as group, and we don't need to capture anything, I replaced them with the non-grouping version (?:..).
And in this case we can do the global search-and-replace once on the giant string instead of individually on each pattern. So the result is the code above:
const allowedRegexStr = '^(?:' +
allowedIPpatterns.
join('|').
replace(/\./g, '\\.').
replace(/X/g, '\\d+') +
')$';
That's still just a string; we have to turn it into an actual RegExp object to do the matching:
const allowedRegexp = new RegExp(allowedRegexStr);
As written, this doesn't filter out illegal IPs - for instance, 10.1234.5678.9012 would match the first pattern. If you want to limit the individual byte values to the decimal range 0-255, you can use a more complicated regex than \d+, like this:
(?:\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])
That matches "any one or two digits, or '1' followed by any two digits, or '2' followed by any of '0' through '4' followed by any digit, or '25' followed by any of '0' through '5'". Replacing the \d with that turns the full string-munging expression into this:
const allowedRegexStr = '^(?:' +
allowedIPpatterns.
join('|').
replace(/\./g, '\\.').
replace(/X/g, '(?:\\d{1,2}|1\\d{2}|2[0-4]\\d|25[0-5])') +
')$';
And makes the actual regex look much more unwieldy:
^(?:10\.(?:\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])\.(?:\d{1,2}|1\d{2}|2[0-4]\d|25[0-5]).(?:\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])|12\.122\....
but you don't have to look at it, just match against it. :)

You could do it in regex, but it's not going to be pretty, especially since JavaScript doesn't even support verbose regexes, which means that it has to be one humongous line of regex without any comments. Furthermore, regexes are ill-suited for matching ranges of numbers. I suspect that there are better tools for dealing with this.
Well, OK, here goes (for the samples you provided):
var myregexp = /\b(?:74\.23\.211\.92|(?:12\.(?:122|211)|64\.23)\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])|(?:10|64)\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9]))\b/g;
As a verbose ("readable") regex:
\b # start of number
(?: # Either match...
74\.23\.211\.92 # an explicit address
| # or
(?: # an address that starts with
12\.(?:122|211) # 12.122 or 12.211
| # or
64\.23 # 64.23
)
\. # .
(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\. # followed by 0..255 and a dot
(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9]) # followed by 0..255
| # or
(?:10|64) # match 10 or 64
\. # .
(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\. # followed by 0..255 and a dot
(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\. # followed by 0..255 and a dot
(?:25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9]) # followed by 0..255
)
\b # end of number

/^(X|\d{1,3})(\.(X|\d{1,3})){3}$/ should do it.

If you don't actually need to match the "X" character you could use this:
\b(?:\d{1,3}\.){3}\d{1,3}\b
Otherwise I would use the solution cebarrett provided.

I'm not entirely sure of what you're trying to achieve here (doesn't look anyone else is either).
However, if it's validation, then here's a solution to validate an IP address that doesn't use RegEx. First, split the input string at the dot. Then using parseInt on the number, make sure it isn't higher than 255.
function ipValidator(ipAddress) {
var ipSegments = ipAddress.split('.');
for(var i=0;i<ipSegments.length;i++)
{
if(parseInt(ipSegments[i]) > 255){
return 'fail';
}
}
return 'match';
}
Running the following returns 'match':
document.write(ipValidator('10.255.255.125'));
Whereas this will return 'fail':
document.write(ipValidator('10.255.256.125'));
Here's a noted version in a jsfiddle with some examples, http://jsfiddle.net/VGp2p/2/