We Keep Coding

Regex match url with params to specific pattern but not query string

My regex pattern:
const pattern = /^\/(test|foo|bar\/baz|en|ppp){1}/i;
const mat = pattern.exec(myURL);
I want to match:
www.mysite.com/bar/baz/myParam/...anything here
but not
www.mysite.com/bar/baz/?uid=100/..
myParam can be any string with or without dashes but only after that anything else can occur like query strings but not immediately after baz.
Tried
/^\/(test|foo|bar\/baz\/[^/?]*|en|ppp){1}/i;
Nothing works.

This, I believe, is what you are asking for:
const myURL = "www.mysite.com/bar/baz/myParam/";
const myURL2 = "www.mysite.com/bar/baz/?uid=100";
const regex = /\/[^\?]\w+/gm;
console.log('with params', myURL.match(regex));
console.log('with queryParams', myURL2.match(regex))
You can test this and play further in Regex101. Even more, if you use that page, it tells you what does what in the regex string.
If it's not what you were asking for, there was another question related to yours, without regex: Here it is

For the 2 example strings, you might use
^[^\/]+\/bar\/baz\/[\w-]+\/.*$
Regex demo
If you want to use the alternations as well, it might look like
^[^\/]+\/(?:test|foo|bar)\/(?:baz|en|ppp)\/[\w-]+\/.*$
^ Start of string
[^\/]+ Match 1+ times any char except a /
\/ Match /
(?:test|foo|bar) Match 1 of the options
\/ Match /
(?:baz|en|ppp) Match 1 of the options
\/ Match /
[\w-]+ Match 1+ times a word char or -
\/ Match /
.* Match 0+ occurrences of any char except a newline
$ End of string
Regex demo

Using a negative lookahead or lookbehind will solve your problem. There are 2 options not clear from the question:
?uid=100 is not allowed after the starting part /bar/baz, so www.mysite.com/test/bar/baz?uid=100 should be valid.
?uid=100 is not allowed anywhere in the string following /bar/baz, which means that www.mysite.com/test/bar/baz/?uid=100 is invalid as well.
Option 1
In short:
\/(test|foo|bar\/baz(?!\/?\?)|en|ppp)(\/[-\w?=]+)*\/?
Explanation of the important parts:
| # OR
bar # 'bar' followed by
\/ # '/' followed by
baz # 'baz'
(?! # (negative lookahead) so, **not** followed by
\/? # 0 or 1 times '/'
\? # '?'
) # END negative lookahead
and
( # START group
\/ # '/'
[-\w?=]+ # any word char, or '-','?','='
)* # END group, occurrence 0 or more times
\/? # optional '/'
Examples Option 1
You can make the lookahead even more specific with something like (?!\/?\?\w+=\w+) to make explicit that ?a=b is not allowed, but that's up to you.
Option 2
To make explicit that ?a=b is not allowed anywhere we can use negative lookbehind. Let's first find a solution for not allowing* bar/baz preceding the ?a=b.
Shorthand:
(?<!bar\/baz\/?)\?\w+=\w+
Explanation:
(?<! # Negative lookbehind: do **not** match preceding
bar\/baz # 'bar/baz'
\/? # optional '/'
)
\? # match '?'
\w+=\w+ # match e.g. 'a=b'
Let's make this part of the complete regex:
\/(test|foo|en|ppp|bar\/baz)(\/?((?<!bar\/baz\/?)\?\w+=\w+|[-\w]+))*\/?$
Explanation:
\/ # match '/'
(test|foo|en|ppp|bar\/baz) # start with 'test', 'foo', 'en', 'ppp', 'bar/baz'
(\/? # optional '/'
((?<!bar\/baz\/?)\?\w+=\w+ # match 'a=b', with negative lookbehind (see above)
| # OR
[-\w]+) # 1 or more word chars or '-'
)* # repeat 0 or more times
\/? # optional match for closing '/'
$ # end anchor
Examples Option 2

regex to only allow an input to 3 decimal places with 0.001 being the smallest number possible, not 0

I am trying to write a regex to allow a user enter a positive number and to 3 decimal places. My regex looks like this, however, it isn't working as I would like.
/\d*[1-9](\.\d{0,3})?/
This allows the user to enter 1.000 as the smallest number, however, it doesn't allow a user to enter 0.001 which should be the smallest number possible to enter into the input.
Does anyone know what the regex should be to solve this?

Your code has another issue where it can not match 10 since you are not allowing the ones place to be 0.
You need to use some or statements
const re = /(^([1-9]|\d{2,})(\.\d{0,3})?|0\.\d{0,2}[1-9])$/
const tests = ["0.001", "0.1", "0","0.0", "0.000","10.001", "10","11","1"]
tests.forEach(n => console.log(n, re.test(n)))

const re = /^(?!0+(?:\.0+)?$)\d+(?:\.\d+)?$/
const tests = ["0.001", "0.1", "0","0.0", "0.000","10.001", "10","11","1","1.22","1.222"]
tests.forEach(n => console.log(n, re.test(n)))
Explanation:
^ # beginning of string
(?! # negative lookahead, make sure we haven't after:
0+ # 1 or more zero
(?: # start non capture group
\. # a dot
0+ # 1 or more zero
)? # end group, optional
$ # end of string
) # end lookahead
\d+ # 1 or more digits
(?: # start non capture group
\. # a dot
\d+ # 1 or more digits
)? # end group, optionnal
$ # end of string

Personally I would just check for 0 and make the regex a lot simpler, but here is a solution, where the required decimal places can be adjusted by changing {1,3}.
The jist of this regex is that we allow any number greater than two digits , then allow only 1-9 for one digit, then optionally require up to 1 decimal with 1-3 digits afterwards.
const r = /^((([0-9]{2,}){1}|[1-9]{1})(\.[0-9]{1,3}){0,1})$/;
const tests = ['1','2','0','1.001','1.001.1','999.001','9.01','9.0100','abc'];
tests.forEach(t=>console.log(t,r.test(t)));

Another option is to use a negative lookahead to assert from the start of the string what is on the right is neither a dot or zero repeated until the end of the string:
^(?![0.]+$)\d+(?:\.\d{1,3})?$
See a Regex demo
Explanation
^ Start of the string
(?![0.]+$) Negative lookahead to assert what is on the right is not what is listed in the character class repeated 1+ times until the end of the string
\d+ Match 1+ times a digit
(?:\.\d{1,3})? Optional non capturing group which matches a dot and 1+ times a digit
$ End of the string

const tests = ["0.001", "0.1", "0","0.0", "0.000","10.001", "10","11","1","1.22","1.222"]
tests.forEach(n => console.log(parseFloat(n) >= 0.001))
I really think this is being overthought.

The answer is here.
([1-9]\.[0-9][0-9][0-9]|[0]\.[1-9][0-9][0-9]|[0]\.[0][1-9][0-9]|[0]\.[0][0][1-9])
This should match 0.001~9.999

Regular expression capture with optional trailing underscore and number

I'm trying to find a regular expression that will match the base string without the optional trailing number (_123). e.g.:
lorem_ipsum_test1_123 -> capture lorem_ipsum_test1
lorem_ipsum_test2 -> capture lorem_ipsum_test2
I tried using the following expression, but it would only work when there is a trailing _number.
/(.+)(?>_[0-9]+)/
/(.+)(?>_[0-9]+)?/
Similarly, adding the ? (zero or more) quantifier only worked when there is no trailing _number, otherwise, the trailing _number would just be part of the first capture.
Any suggestions?

You may use the following expression:
^(?:[^_]+_)+(?!\d+$)[^_]+
^ Anchor beginning of string.
(?:[^_]+_)+ Repeated non capturing group. Negated character set for anything other than a _, followed by a _.
(?!\d+$) Negative lookahead for digits at the end of the string.
[^_]+ Negated character set for anything other than a _.
Regex demo here.
Please note that the \n in the character sets in the Regex demo are only for demonstration purposes, and should by all means be removed when using as a pattern in Javascript.
Javascript demo:
var myString = "lorem_ipsum_test1_123";
var myRegexp = /^(?:[^_]+_)+(?!\d+$)[^_]+/g;
var match = myRegexp.exec(myString);
console.log(match[0]);
var myString = "lorem_ipsum_test2"
var myRegexp = /^(?:[^_]+_)+(?!\d+$)[^_]+/g;
var match = myRegexp.exec(myString);
console.log(match[0]);

You might match any character and use a negative lookahead that asserts that what follows is not an underscore, one or more digits and the end of the string:
^(?:(?!_\d+$).)*
Explanation
^ Assert start of the string
(?: Non capturing group
(?! Negative lookahead to assert what is on the right side is not
_\d+$Match an underscore, one or more digits and assert end of the string
.) Match any character and close negative lookahead
)* Close non capturing group and repeat zero or more times
Regex demo
const strings = [
"lorem_ipsum_test1_123",
"lorem_ipsum_test2"
];
let pattern = /^(?:(?!_\d+$).)*/;
strings.forEach((s) => {
console.log(s + " ==> " + s.match(pattern)[0]);
});

You are asking for
/^(.*?)(?:_\d+)?$/
See the regex demo. The point here is that the first dot pattern must be non-greedy and the _\d+ should be wrapped with an optional non-capturing group and the whole pattern (especially the end) must be enclosed with anchors.
Details
^ - start of string
(.*?) - Capturing group 1: any zero or more chars other than line break chars, as few as possible due to the non-greedy ("lazy") quantifier *?
(?:_\d+)? - an optional non-capturing group matching 1 or 0 occurrences of _ and then 1+ digits
$ - end of string.
However, it seems easier to use a mere replacing approach,
s = s.replace(/_\d+$/, '')
If the string ends with _ and 1+ digits, the substring will get removed, else, the string will not change.
See this regex demo.

Try to check if the string contains the trailing number. If it does you get only the other part. Otherwise you get the whole string.
var str = "lorem_ipsum_test1_123"
if(/_[0-9]+$/.test(str)) {
console.log(str.match(/(.+)(?=_[0-9]+)/g))
} else {
console.log(str)
}
Or, a lot more concise:
str = str.replace(/_[0-9]+$/g, "")

How do I combine these two regular expressions into one?

I'm writing a rudimentary lexer using regular expressions in JavaScript and I have two regular expressions (one for single quoted strings and one for double quoted strings) which I wish to combine into one. These are my two regular expressions (I added the ^ and $ characters for testing purposes):
var singleQuotedString = /^'(?:[^'\\]|\\'|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*'$/gi;
var doubleQuotedString = /^"(?:[^"\\]|\\"|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*"$/gi;
Now I tried to combine them into a single regular expression as follows:
var string = /^(["'])(?:[^\1\\]|\\\1|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*\1$/gi;
However when I test the input "Hello"World!" it returns true instead of false:
alert(string.test('"Hello"World!"')); //should return false as a double quoted string must escape double quote characters
I figured that the problem is in [^\1\\] which should match any character besides matching group \1 (which is either a single or a double quote - the delimiter of the string) and \\ (which is the backslash character).
The regular expression correctly filters out backslashes and matches the delimiters, but it doesn't filter out the delimiter within the string. Any help will be greatly appreciated. Note that I referred to Crockford's railroad diagrams to write the regular expressions.

You can't refer to a matched group inside a character class: (['"])[^\1\\]. Try something like this instead:
(['"])((?!\1|\\).|\\[bnfrt]|\\u[a-fA-F\d]{4}|\\\1)*\1
(you'll need to add some more escapes, but you get my drift...)
A quick explanation:
(['"]) # match a single or double quote and store it in group 1
( # start group 2
(?!\1|\\). # if group 1 or a backslash isn't ahead, match any non-line break char
| # OR
\\[bnfrt] # match an escape sequence
| # OR
\\u[a-fA-F\d]{4} # match a Unicode escape
| # OR
\\\1 # match an escaped quote
)* # close group 2 and repeat it zero or more times
\1 # match whatever group 1 matched

This should work too (raw regex).
If speed is a factor, this is the 'unrolled' method, said to be the fastest for this kind of thing.
(['"])(?:(?!\\|\1).)*(?:\\(?:[\/bfnrt]|u[0-9A-F]{4}|\1)(?:(?!\\|\1).)*)*/1
Expanded
(['"]) # Capture a quote
(?:
(?!\\|\1). # As many non-escape and non-quote chars as possible
)*
(?:
\\ # escape plus,
(?:
[\/bfnrt] # /,b,f,n,r,t or u[a-9A-f]{4} or captured quote
| u[0-9A-F]{4}
| \1
)
(?:
(?!\\|\1). # As many non-escape and non-quote chars as possible
)*
)*
/1 # Captured quote

Well, you can always just create a larger regex by just using the alternation operator on the smaller regexes
/(?:single-quoted-regex)|(?:double-quoted-regex)/
Or explicitly:
var string = /(?:^'(?:[^'\\]|\\'|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*'$)|(?:^"(?:[^"\\]|\\"|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*"$)/gi;
Finally, if you want to avoid the code duplication, you can build up this regex dynamically, using the new Regex constructor.
var quoted_string = function(delimiter){
return ('^' + delimiter + '(?:[^' + delimiter + '\\]|\\' + delimiter + '|\\\\|\\\/|\\b|\\f|\\n|\\r|\\t|\\u[0-9A-F]{4})*' + delimiter + '$').replace(/\\/g, '\\\\');
//in the general case you could consider using a regex excaping function to avoid backslash hell.
};
var string = new RegExp( '(?:' + quoted_string("'") + ')|(?:' + quoted_string('"') + ')' , 'gi' );

Regular expression negative match

I can't seem to figure out how to compose a regular expression (used in Javascript) that does the following:
Match all strings where the characters after the 4th character do not contain "GP".
Some example strings:
EDAR - match!
EDARGP - no match
EDARDTGPRI - no match
ECMRNL - match
I'd love some help here...

Use zero-width assertions:
if (subject.match(/^.{4}(?!.*GP)/)) {
// Successful match
}
Explanation:
"
^ # Assert position at the beginning of the string
. # Match any single character that is not a line break character
{4} # Exactly 4 times
(?! # Assert that it is impossible to match the regex below starting at this position (negative lookahead)
. # Match any single character that is not a line break character
* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
GP # Match the characters “GP” literally
)
"

You can use what's called a negative lookahead assertion here. It looks into the string ahead of the location and matches only if the pattern contained is /not/ found. Here is an example regular expression:
/^.{4}(?!.*GP)/
This matches only if, after the first four characters, the string GP is not found.

could do something like this:
var str = "EDARDTGPRI";
var test = !(/GP/.test(str.substr(4)));
test will return true for matches and false for non.