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JavaScript Regex split at first letter?

Since many cases using Regex, differs from case to case, depending on what format your string is in, I'm having a hard time finding a solution to my problem.
I have an array containing strings in the format, as an example:
"XX:XX - XX:XX Algorithm and Data Structures"
Where "XX:XX - XX:XX" is timespan for a lecture, and X being a number.
I'm new to Regex and trying to split the string at the first letter occurring, like so:
let str = "08:15 - 12:50 Algorithm and Data Structures";
let re = //Some regex expression
let result = str.split(re); // Output: ["08:15 - 12:50", "Algorithm and Data Structures"]
I'm thinking it should be something like /[a-Z]/ but I'm not sure at all...
Thanks in advance!

The easiest way is probably to "mark" where you want to split and then split:
const str = '12 34 abcde 45 abcde'.replace(/^([^a-z]+)([a-z])/i, '$1,$2');
// '12 34 ,abcde 45 abcde'
str.split(',')
// [ '12 34 ', 'abcde 45 abcde' ]
This finds the place where the string starts, has a bunch of non a-z characters, then has an a-z characters, and puts a comma right in-between. Then you split by the comma.
You can also split directly with a positive look ahead but it might make the regex a bit less readable.

console.log(
"08:15 - 12:50 Algorithm and Data Structures".split(/ ([A-Za-z].*)/).filter(Boolean)
)
or, if it's really always XX:XX - XX:XX, easier to just do:
const splitTimeAndCourse = (input) => {
return [
input.slice(0, "XX:XX - XX:XX".length),
input.slice("XX:XX - XX:XX".length + 1)
]
}
console.log(splitTimeAndCourse("08:15 - 12:50 Algorithm and Data Structures"))

If you have a fixed length of the string where the time is, you can use this regex for example
(^.{0,13})(.*)
Check this here https://regex101.com/r/ANMHy5/1

I know you asked about regex in particular, but here is a way to this without regex...
Provided your time span is always at the beginning of your string and will always be formatted with white space between the numbers as XX:XX - XX:XX. You could use a function that splits the string at the white space and reconstructs the first three indexed strings into one chunk, the time span, and the last remaining strings into a second chunk, the lecture title. Then return the two chunks as an array.
let str = "08:15 - 12:50 Algorithm and Data Structures";
const splitString = (str) => {
// split the string at the white spaces
const strings = str.split(' ')
// define variables
let lecture = '',
timespan = '';
// loop over the strings
strings.forEach((str, i) => {
// structure the timespan
timespan = `${strings[0]} ${strings[1]} ${strings[2]}`;
// conditional to get the remaining strings and concatenate them into a new string
i > 2 && i < strings.length?lecture += `${str} `: '';
})
// place them into an array and remove white space from end of second string
return [timespan, lecture.trimEnd()]
}
console.log(splitString(str))

For that format, you might also use 2 capture groups instead of using split.
^(\d{1,2}:\d{1,2}\s*-\s*\d{1,2}:\d{1,2})\s+([A-Za-z].*)
The pattern matches:
^ Start of string
(\d{1,2}:\d{1,2}\s*-\s*\d{1,2}:\d{1,2}) Capture group 1, match a timespan like pattern
\s+ Match 1+ whitspac chars
([A-Za-z].*) Capture group 2, start with a char A-Za-z and match the rest of the line.
Regex demo
let str = "08:15 - 12:50 Algorithm and Data Structures";
let regex = /^(\d{1,2}:\d{1,2}\s*-\s*\d{1,2}:\d{1,2})\s+([A-Za-z].*)/;
let [, ...groups] = str.match(regex);
console.log(groups);
Another option using split might be asserting not any chars a-zA-Z to the left from the start of the string using a lookbehind (see this link for the support), match 1+ whitespace chars and asserting a char a-zA-Z to the right.
(?<=^[^a-zA-Z]+)\s+(?=[A-Za-z])
Regex demo
let str = "08:15 - 12:50 Algorithm and Data Structures";
let regex = /(?<=^[^a-zA-Z]+)\s+(?=[A-Za-z])/;
console.log(str.split(regex))

Javascript replace regex to accept only numbers, including negative ones, two decimals, replace 0s in the beginning, except if number is 0

The question became a bit long, but it explains the expected behaviour.
let regex = undefined;
const format = (string) => string.replace(regex, '');
format('0')
//0
format('00')
//0
format('02')
//2
format('-03')
//-3
format('023.2323')
//23.23
format('00023.2.3.2.3')
//23.23
In the above example you can see the expected results in comments.
To summarize. I'm looking for a regex not for test, for replace which formats a string:
removes 0s from the beginning if it's followed by any numbers
allows decimal digits, but just 2
allows negative numbers
allows decimal points, but just one (followed by min 1, max 2 decimal digits)
The last one is a bit difficult to handle as the user can't enter period at the same time, I'll have two formatter functions, one will be the input in the input field, and one for the closest valid value at the moment (for example '2.' will show '2.' in the input field, but the handler will receive the value '2').
If not big favour, I'd like to see explanation of the solution, why it works, and what's the purpose of which part.
Right now I'm having string.replace(/[^\d]+(\.\[^\d{1,2}])+|^0+(?!$)/g, ''), but it doesn't fulfill all the requirements.

You may use this code:
const arr = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3']
var narr = []
// to remove leading zeroes
const re1 = /^([+-]?)0+?(?=\d)/
// to remove multiple decimals
const re2 = /^([+-]?\d*\.\d+)\.(\d+).*/
arr.forEach( el => {
el = el.replace(re1, '$1').replace(re2, '$1$2')
if (el.indexOf('.') >= 0)
el = Number(el).toFixed(2)
narr.push(el)
})
console.log(narr)
//=> ["0", "0", "2", "-3", "23.23", "23.23"]

If you aren't bound to the String#replace method, you can try this regex:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
Inspect on regex101.com
It collects the parts of the number into capturing groups, as follows:
Sign: the sign of the number, +, - or undefined
Integer: the integer part of the number, without leading zeros
Decimal: the decimal part of the number, undefined if absent
This regex won't match if more then 2 decimal places present. To strip it instead, use this:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
Inspect on regex101.com
To format a number using one of the above, you can use something like:
let regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
const format = string => {
try{
const [, sign, integer, decimal = ''] = string.match(regex)
return `${(sign !== '-' ? '' : '-')}${integer}${(decimal && `.${decimal}`)}`
}catch(e){
//Invalid format, do something
return
}
}
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//undefined (invalid format)
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)
//Using the 2nd regex
regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//23.23
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)

Another option is to use pattern with 3 capturing groups. In the replacement, use all 3 groups "$1$2$3"
If the string after the replacement is empty, return a single zero.
If the string is not empty, concat group 1, group 2 and group 3 where for group 3, remove all the dots except for the first one to keep it for the decimal and take the first 3 characters (which is the dot and 2 digits)
^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$
In parts
^ Start of string
( Capture group 1
[-+]? Match an optional + or -
) Close group
0* Match 0+ times a zero
( Capture group 2
[1-9]\d* Match a digit 1-9 followed by optional digits 0-9
) Close group
( Capture group 3
(?:\.\d+)* Repeat 0+ times matching a dot and a digit
) Close group
| Or
0+ Match 1+ times a zero
$ End of string
Regex demo
const strings = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3', '00001234', '+0000100005.0001']
let pattern = /^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$/;
let format = s => {
s = s.replace(pattern, "$1$2$3");
return s === "" ? '0' : s.replace(pattern, (_, g1, g2, g3) =>
g1 + g2 + g3.replace(/(?!^)\./g, '').substring(0, 3)
);
};
strings.forEach(s => console.log(format(s)));

Parse query parameters with regexp

I need to parse the url /domain.com?filter[a.b.c]=value1&filter[a.b.d]=value2
and get 2 groups: 'a.b.c' and 'a.b.d'.
I try to parse with regexp [\?&]filter\[(.+\..+)+\]= but the result is 'a.b.c]=value1&filter[a.b.d'. How can I specify to search for the 1st occurrence?

You may use
/[?&]filter\[([^\].]+\.[^\]]+)]=/g
See the regex demo
Details
[?&] - a ? or &
filter\[ - a filter[ substring
([^\].]+\.[^\]]+) - Capturing group 1:
[^\].]+ - 1 or more chars other than ] and .
\. - a dot
[^\]]+ - 1 or more chars other than ]
]= - a ]= substring
JS demo:
var s = '/domain.com?filter[a.b.c]=value1&filter[a.b.d]=value2';
var rx = /[?&]filter\[([^\].]+\.[^\]]+)]=/g;
var m, res=[];
while(m=rx.exec(s)) {
res.push(m[1]);
}
console.log(res);
Note that in case & is never present as part of the query param value, you may add it to the negated character classes, [^\].]+ => [^\]&.]+, to make sure the regex does not overmatch across param values.
Since you need to extract text inside outer square brackets that may contain consecutive [...] substrings with at least 1 dot inside one of them, you may use a simpler regex with a bit more code:
var strs = ['/domain.com?filter[a.b.c]=value1&filter[a.b.d]=value2',
'/domain.com?filter[a.b.c]=value1&filter[a.b.d]=value2&filter[a][b.e]=value3',
'/domain.com?filter[a.b.c]=value1&filter[b][a.b.d][d]=value2&filter[a][b.e]=value3'];
var rx = /[?&]filter((?:\[[^\][]*])+)=/g;
for (var s of strs) {
var m, res=[];
console.log(s);
while(m=rx.exec(s)) {
if (m[1].indexOf('.') > -1) {
res.push(m[1].substring(1,m[1].length-1));
}
}
console.log(res);
console.log("--- NEXT STRING ----");
}

(?<=[\?&]filter\[)([^\]]+\.[^\]]+)+(?!>\]=)
This will give you only the groups you mentioned (a.b.c and a.b.d)
This part (?<=[\?&]filter\[) says recognise but don't capture [?&]filter before what you want and this part (?!>\]=) says recognise but don't capture after ] after what you want.
[^\]] this captures everything that isn't a square bracket

regex to extract numbers starting from second symbol

Sorry for one more to the tons of regexp questions but I can't find anything similar to my needs. I want to output the string which can contain number or letter 'A' as the first symbol and numbers only on other positions. Input is any string, for example:
---INPUT--- -OUTPUT-
A123asdf456 -> A123456
0qw#$56-398 -> 056398
B12376B6f90 -> 12376690
12A12345BCt -> 1212345
What I tried is replace(/[^A\d]/g, '') (I use JS), which almost does the job except the case when there's A in the middle of the string. I tried to use ^ anchor but then the pattern doesn't match other numbers in the string. Not sure what is easier - extract matching characters or remove unmatching.

I think you can do it like this using a negative lookahead and then replace with an empty string.
In an non capturing group (?:, use a negative lookahad (?! to assert that what follows is not the beginning of the string followed by ^A or a digit \d. If that is the case, match any character .
(?:(?!^A|\d).)+
var pattern = /(?:(?!^A|\d).)+/g;
var strings = [
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
for (var i = 0; i < strings.length; i++) {
console.log(strings[i] + " ==> " + strings[i].replace(pattern, ""));
}

You can match and capture desired and undesired characters within two different sides of an alternation, then replace those undesired with nothing:
^(A)|\D
JS code:
var inputStrings = [
"A-123asdf456",
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
console.log(
inputStrings.map(v => v.replace(/^(A)|\D/g, "$1"))
);

You can use the following regex : /(^A)?\d+/g
var arr = ['A123asdf456','0qw#$56-398','B12376B6f90','12A12345BCt', 'A-123asdf456'],
result = arr.map(s => s.match(/(^A|\d)/g).join(''));
console.log(result);

javascript regular expression test for 6 digit numbers only. comma seperated

and so this must pass:
454555, 939999 , 019999 ,727663
its for a user entering 6 digit invoice numbers. it should fail if a number is 5 or 7 digit and not 6. so 1234567, 123456 should fail, as one set is more than 6 numbers.
So far I have :
[0-9]{6}(\s*,*,\s*[0-9]{6})*
which only draw back is that it accepts 7 or more digit numbers. cant figure out if its even possible at this point to do both, test for 6 digits separated by a comma and one or more space, and all the digits have to be only 6 digits and fail if one is not.
any help appreciated. regular expressions are not my forte.
thanks
Norman

You can write it using regex like the function below.
const isPassword = (password: string) => /^\d{6}$/gm.test(password);
And here is an example test file below.
test('should recognize a valid password', () => {
expect(isPassword('123456')).toBe(true);
expect(isPassword('000000')).toBe(true);
});
test('should recognize an invalid password', () => {
expect(isPassword('asdasda1234')).toBe(false);
expect(isPassword('1234567')).toBe(false);
expect(isPassword('a123456a')).toBe(false);
expect(isPassword('11.11.11')).toBe(false);
expect(isPassword('aaaaaa')).toBe(false);
expect(isPassword('eeeeee')).toBe(false);
expect(isPassword('......')).toBe(false);
expect(isPassword('werwerwerwr')).toBe(false);
});

In order to validate the full string you can use this regex.
^(\s*\d{6}\s*)(,\s*\d{6}\s*)*,?\s*$
It works with six digits only, and you have to enter at least one 6 digit number.
It also works if you have a trailing comma with whitespaces.

It's accepting more than six digit numbers because you're not anchoring the text, and for some odd reason you're optionally repeating the comma. Try something like this:
^[0-9]{6}(?:\s*,\s*[0-9]{6})*$
Also note that [0-9] is equivalent to \d, so this can be rewritten more concisely as:
^\d{6}(?:\s*,\s*\d{6})*$

Your regex does not match 7 digits in a row, but it also doesn't enforce that it matches the whole string. It just has to match some substring in the string, so it would also match each of these:
"1234512345612345612345"
"NaNaNaN 123456, 123456 BOOO!"
"!##$%^&*({123456})*&^%$##!"
Just add the start of string (^) and end of string ($) anchors to enforce that the whole string matches and it will work correctly:
^[0-9]{6}(\s*,*,\s*[0-9]{6})*$
Also note that ,*, could be shortened to ,+, and if you only want one comma in a row, just use ,, not ,* or ,+.
You can also replace [0-9] with \d:
^\d{6}(\s*,\s*\d{6})*$

Using only regex:
var commaSeparatedSixDigits = /^(?:\d{6}\s*,\s*)*\d{6}$/;
if (myInput.test(commaSeparatedSixDigits)) console.log( "Is good!" );
This says:
^ - Starting at the beginning of the string
(?:…)* - Find zero or more of the following:
\d{6} - six digits
\s* - maybe some whitespace
, - a literal comma
\s* - maybe some whitespace
\d{6} - Followed by six digits
$ - Followed by the end of the string
Alternatively:
var commaSeparatedSixDigits = /^\s*\d{6}(?:\s*,\s*\d{6})*\s*$/;
I leave it as an exercise to you to decipher what's different about this.
Using JavaScript + regex:
function isOnlyCommaSeparatedSixDigitNumbers( str ){
var parts = srt.split(/\s*,\s*/);
for (var i=parts.length;i--;){
// Ensure that each part is exactly six digit characters
if (! /^\d{6}$/.test(parts[i])) return false;
}
return true;
}

I see a lot of complication here. Sounds to me like what you want is pretty simple:
/^(\d{6},)*\d{6}$/
Then we account for whitespace:
/^\s*(\d{6}\s*,\s*)*\d{6}\s*$/
But as others have noted, this is actually quite simple in JavaScript without using regex:
function check(input) {
var parts = input.split(',');
for (var i = 0, n = parts.length; i < n; i++) {
if (isNaN(+parts[i].trim())) {
return false;
}
}
return true;
}
Tested in the Chrome JavaScript console.

There isn;t any real need for a regexp. Limit the input to only 6 characters, only accept numbers and ensure that the input has 6 digits (not show here). So you would need:
HTML
<input type='text' name='invoice' size='10' maxlength='6' value='' onkeypress='evNumersOnly(event);'>
JavaScript
<script>
function evNumbersOnly( evt ) {
//--- only accepts numbers
//--- this handles incompatabilities between browsers
var theEvent = evt || window.event;
//--- this handles incompatabilities between browsers
var key = theEvent.keyCode || theEvent.which;
//--- convert key number to a letter
key = String.fromCharCode( key );
var regex = /[0-9]/; // Allowable characters 0-9.+-,
if( !regex.test(key) ) {
theEvent.returnValue = false;
//--- this prevents the character from being displayed
if(theEvent.preventDefault) theEvent.preventDefault();
}
}
</script>