I'm trying to create a regex that will select the numbers/numbers with commas(if easier, can trim commas later) that do not have a parentheses after and not the numbers inside the parentheses should not be selected either.
Used with the JavaScript's String.match method
Example strings
9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
What i have so far:
/((^\d+[^\(])|(,\d+,)|(,*\d+$))/gm
I tried this in regex101 and underlined the numbers i would like to match and x on the one that should not.
You could start with a substitution to remove all the unwanted parts:
/\d*\(.*?\),?//gm
Demo
This leaves you with
5,10
10,2,5,
10,7,2,4
which makes the matching pretty straight forward:
/(\d+)/gm
If you want it as a single match expression you could use a negative lookbehind:
/(?<!\([\d,]*)(\d+)(?:,|$)/gm
Demo - and here's the same matching expression as a runnable javascript (skeleton code borrowed from Wiktor's answer):
const text = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4`;
const matches = Array.from(text.matchAll(/(?<!\([\d,]*)(\d+)(?:,|$)/gm), x=>x[1])
console.log(matches);
Here, I'd recommend the so-called "best regex trick ever": just match what you do not need (negative contexts) and then match and capture what you need, and grab the captured items only.
If you want to match integer numbers that are not matched with \d+\([^()]*\) pattern (a number followed with a parenthetical substring), you can match this pattern or match and capture the \d+, one or more digit matching pattern, and then simply grab Group 1 values from matches:
const text = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4`;
const matches = Array.from(text.matchAll(/\d+\([^()]*\)|(\d+)/g), x=> x[1] ?? "").filter(Boolean)
console.log(matches);
Details:
text.matchAll(/\d+\([^()]*\)|(\d+)/g) - matches one or more digits (\d+) + ( (with \() + any zero or more chars other than ( and ) (with [^()]*) + \) (see \)), or (|) one or more digits captured into Group 1 ((\d+))
Array.from(..., x=> x[1] ?? "") - gets Group 1 value, or, if not assigned, just adds an empty string
.filter(Boolean) - removes empty strings.
Using several replacement regexes
var textA = `9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
`
console.log('A', textA)
var textB = textA.replace(/\(.*?\),?/g, ';')
console.log('B', textB)
var textC = textB.replace(/^\d+|\d+$|\d*;\d*/gm, '')
console.log('C', textC)
var textD = textC.replace(/,+/g, ' ').trim(',')
console.log('D', textD)
With a loop
Here is a solution which splits the lines on comma and loops over the pieces:
var inside = false;
var result = [];
`9(296,178),5,3(123),10
10,9(296,178),2,5,3(123),3(124,125)
10,7,5(296,293,444,1255),3(218),2,4
`.split("\n").map(line => {
let pieceArray = line.split(",")
pieceArray.forEach((piece, k) => {
if (piece.includes('(')) {
inside = true
} else if (piece.includes(')')) {
inside = false
} else if (!inside && k > 0 && k < pieceArray.length-1 && !pieceArray[k-1].includes(')')) {
result.push(piece)
}
})
})
console.log(result)
It does print the expected result: ["5", "7"]
Related
I’m struggling with this simple regex that is not working correctly in Safari:
(?<=\?.*)\?
It should match each ?, except of the first one.
I know that lookbehind is not working on Safari yet, but I need to find some workaround for it. Any suggestions?
You can use an alternation capture until the first occurrence of the question mark. Use that group again in the replacement to leave it unmodified.
In the second part of the alternation, match a questionmark to be replaced.
const regex = /^([^?]*\?)|\?/g;
const s = "test ? test ? test ?? test /";
console.log(s.replace(regex, (m, g1) => g1 ? g1 : "[REPLACE]"));
There are always alternatives to lookbehinds.
In this case, all you need to do is replace all instances of a character (sequence), except the first.
The .replace method accepts a function as the second argument.
That function receives the full match, each capture group match (if any), the offset of the match, and a few other things as parameters.
.indexOf can report the first offset of a match.
Alternatively, .search can also report the first offset of a match, but works with regexes.
The two offsets can be compared inside the function:
const yourString = "Hello? World? What? Who?",
yourReplacement = "!",
pattern = /\?/g,
patternString = "?",
firstMatchOffsetIndexOf = yourString.indexOf(patternString),
firstMatchOffsetSearch = yourString.search(pattern);
console.log(yourString.replace(pattern, (match, offset) => {
if(offset !== firstMatchOffsetIndexOf){
return yourReplacement;
}
return match;
}));
console.log(yourString.replace(pattern, (match, offset) => {
if(offset !== firstMatchOffsetSearch){
return yourReplacement;
}
return match;
}));
This works for character sequences, too:
const yourString = "Hello. Hello. Hello. Hello.",
yourReplacement = "Hi",
pattern = /Hello/g,
firstOffset = yourString.search(pattern);
console.log(yourString.replace(pattern, (match, offset) => {
if(offset !== firstOffset){
return yourReplacement;
}
return match;
}));
Split and join with
var s = "one ? two ? three ? four"
var l = s.split("?") // Split with ?
var first = l.shift() // Get first item and remove from l
console.log(first + "?" + l.join("<REPLACED>")) // Build the results
I need to parse a string that comes like this:
-38419-indices-foo-7119-attributes-10073-bar
Where there are numbers followed by one or more words all joined by dashes. I need to get this:
[
0 => '38419-indices-foo',
1 => '7119-attributes',
2 => '10073-bar',
]
I had thought of attempting to replace only the dash before a number with a : and then using .split(':') - how would I do this? I don't want to replace the other dashes.
Imo, the pattern is straight-forward:
\d+\D+
To even get rid of the trailing -, you could go for
(\d+\D+)(?:-|$)
Or
\d+(?:(?!-\d|$).)+
You can see it here:
var myString = "-38419-indices-foo-7119-attributes-10073-bar";
var myRegexp = /(\d+\D+)(?:-|$)/g;
var result = [];
match = myRegexp.exec(myString);
while (match != null) {
// matched text: match[0]
// match start: match.index
// capturing group n: match[n]
result.push(match[1]);
match = myRegexp.exec(myString);
}
console.log(result);
// alternative 2
let alternative_results = myString.match(/\d+(?:(?!-\d|$).)+/g);
console.log(alternative_results);
Or a demo on regex101.com.
Logic
lazy matching using quantifier .*?
Regex
.*?((\d+)\D*)(?!-)
https://regex101.com/r/WeTzF0/1
Test string
-38419-indices-foo-7119-attributes-10073-bar-333333-dfdfdfdf-dfdfdfdf-dfdfdfdfdfdf-123232323-dfsdfsfsdfdf
Matches
Further steps
You need to split from the matches and insert into your desired array.
Sorry if the wording is bad. So I'm trying to find out how to pass in a string match of multiple characters long into my dynamic regex expression.
The regex in my else statement works with 1 character being passed in so I'm trying to do the same thing except with multiple characters being passed in the first if statement.
const delimiter = str.slice(0, str.indexOf('\n'));
const strLength = delimiter.length;
if (delimiter[0] === '[' && delimiter.charAt(strLength - 1) === ']') {
const customDelimiter = delimiter.slice(delimiter.indexOf(delimiter[1]), delimiter.indexOf(delimiter.charAt(strLength - 1)));
console.log(customDelimiter) // => '***'
const regex = new RegExp(`,|\\n|\\${customDelimiter}`,'g');
return strArr = str.split(regex).filter(Boolean);
} else {
const firstChar = str.slice(0, 1); // => '*'
const regex = new RegExp(`,|\\n|\\${firstChar}`,'g');
return strArr = str.split(regex).filter(Boolean);
}
So for example I want this string:
'[*]\n11***22***33' to equal 66 b/c it should split it into an array of [11, 22, 33] using the '*' delimiter. I get an error message saying: "SyntaxError: Invalid regular expression: /,|\n|***/: Nothing to repeat".
When you use * as delimeter in your regex, it becomes ,|\\n|\\|\*, which is the correct regex.
It matches ',' or '\n' or a '*' character.
For your string, it matches [***]\n11***22***33.
But when you use *** as a delimiter in your regex, it becomes ,|\\n|\\|\***, which is incorrect. Here it gets two unescaped * at the end. * in regex means 0 or more of the preceding pattern. You cannot have two of them together.
This is a special case because * has a special meaning in regex.
If you would have used any non-regex character, it would work.
A simpler solution would be to use javascript split function to easily get the desired result.
You could first split the string using \n.
let splitStr = str.split('\n');
// This would return ["[***]", "11***22***33"]
and then split the 1st index of the splitStr using the delimeter.
splitStr[1].split('***');
// splitStr[1].split(customDelimiter)
// This would return ["11", "22", "33"]
Using this you wouldn't need to use if or else statement to separate out single character delimiter and multiple character delimiter.
I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring
Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));
Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer
If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}
There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7
Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)
I have situation where I have to split the string in to 3 substrings.
Example
<=Mindate+30 >>> [<=,Mindate,+30]
<=Mindate >>> [<=,Mindate]
>=Maxdate-3 >>> [>=,Maxdate,-3]
==Date >>> [==,Date]
I have created a function that split current array in two
function splitString(str){
var a= /(>=|<=|=|<|>|\!=)|.+/g
return str.match(a);
}
splitString('<=Mindate+45');
Current output ['<=', 'Mindate+45']
Expected output ['<=', 'Mindate', '+45']
Can any one help me on this ?
You just needed to separate the textual part (\w+) from the number part ([-+]\d+):
function splitToThree(input) {
let regex = /(>=|<=|==|<|>|\!=)|\w+|[-+]\d+/g;
let ans = input.match(regex);
console.log(ans);
}
splitToThree("<=Mindate+30");
splitToThree("<=Mindate");
splitToThree(">=Maxdate-3");
splitToThree("==Date");
You can create a pattern that matches and captures all the comparison operators or matches a place before each of the +, /, * and - operators, and use the expression inside a String#split method:
a.split(/([<>!=]=|[=<>])|(?=[-+\/*])/).filter(Boolean)
See the JS demo:
var strs = ['<=Mindate+30', '<=Mindate','>=Maxdate-3','==Date','>=SomeFn-3.45'];
rx = /([<>!=]=|[=<>])|(?=[-+\/*])/;
for (var a of strs) {
var res = a.split(rx).filter(Boolean);
console.log(a, "=>", res);
}
Pattern details
([<>!=]=|[=<>]) - Group 1 (this value will be part of the resulting array): <, >, !, = followed with = or a =, < or > char
| - or
(?=[-+\/*]) - a location that is followed with -, +, / or *.
Note: .filter(Boolean) will remove empty items from the resulting array.