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I need to manipulate drawing of a SVG, so I have attribute "d" values like this:
d = "M561.5402,268.917 C635.622,268.917 304.476,565.985 379.298,565.985"
What I want is to "purify" all the values (to strip the chars from them), to calculate them (for the sake of simplicity, let's say to add 100 to each value), to deconstruct the string, calculate the values inside and then concatenate it all back together so the final result is something like this:
d = "M661.5402,368.917 C735.622,368.917 404.476,665.985 479.298,665.985"
Have in mind that:
some values can start with a character
values are delimited by comma
some values within comma delimiter can be delimited by space
values are decimal
This is my try:
let arr1 = d.split(',');
arr1 = arr1.map(element => {
let arr2 = element.split(' ');
if (arr2.length > 1) {
arr2 = arr2.map(el => {
let startsWithChar = el.match(/\D+/);
if (startsWithChar) {
el = el.replace(/\D/g,'');
}
el = parseFloat(el) + 100;
if (startsWithChar) {
el = startsWithChar[0] + el;
}
})
}
else {
let startsWithChar = element.match(/\D+/);
if (startsWithChar) {
element = element.replace(/\D/g,'');
}
element = parseFloat(element) + 100;
if (startsWithChar) {
element = startsWithChar[0] + element;
}
}
});
d = arr1.join(',');
I tried with regex replace(/\D/g,'') but then it strips the decimal dot from the value also, so I think my solution is full of holes.
Maybe another solution would be to somehow modify directly each of path values/commands, I'm opened to that solution also, but I don't know how.
const s = 'M561.5402,268.917 C635.622,268.917 304.476,565.985 379.298,565.985'
console.log(s.replaceAll(/[\d.]+/g, m=>+m+100))
You might use a pattern to match the format in the string with 2 capture groups.
([ ,]?\b[A-Z]?)(\d+\.\d+)\b
The pattern matches:
( Capture group 1
[ ,]?\b[A-Z]? Match an optional space or comma, a word boundary and an optional uppercase char A-Z
) Close group 1
( Capture group 2
\d+\.\d+ Match 1+ digits, a dot and 1+ digits
) Close group 1
\b A word boundary to prevent a partial word match
Regex demo
First capture the optional delimiter followed by an optional uppercase char in group 1, and the decimal number in group 2.
Then add 100 to the decimal value and join back the 2 group values.
const d = "M561.5402,268.917 C635.622,268.917 304.476,565.985 379.298,565.985";
const regex = /([ ,]?\b[A-Z]?)(\d+\.\d+)\b/g;
const res = Array.from(
d.matchAll(regex), m => m[1] + (+m[2] + 100)
).join('');
console.log(res);
I found this regex which validates Instagram usernames
/^(?!.*\.\.)(?!.*\.$)[^\W][\w.]{0,29}$/gim
What I'm trying to do is to replace all characters which not match my regex
let regex = new RegExp(/^(?!.*\.\.)(?!.*\.$)[^\W][\w.]{0,29}$/gim);
const filteredString = replace(text, regex, '');
I tried to add ?! at the start as a negative lookahead but no luck
Removing all the aprts that don't mach is the same as keeping the matches.
Instead of using replace you can use match and add all the matches to your filteredString, like shown below:
let text = `riegiejeyaranchen
riegie.jeyaranchen
_riegie.jeyaranchen
.riegie
riegie..jeyaranchen
riegie._.jeyaranchen
riegie.
riegie.__`;
let regex = new RegExp(/^(?!.*\.\.)(?!.*\.$)[^\W][\w.]{0,29}$/gim);
let filteredString = '';
text.match(regex).forEach(value =>
{
filteredString += value + '\r\n';
});
console.log(filteredString);
Of course the \r\n is optional (just places one on each line).
Now you get a string where non matches are removed.
Based on the regex /^(?!.*\.\.)(?!.*\.$)[^\W][\w.]{0,29}$/, the name must not have consecutive dots, no leading and trailing dots, and have max 30 word/dot chars. This code cleans up names accordingly:
[
'f',
'foobar',
'_foo.bar',
'_foo..bar',
'_foo...bar',
'_foo.bar.',
'.foo.bar',
'foo<$^*>bar',
'123456789012345678901234567890',
'1234567890123456789012345678901'
].forEach(name => {
let clean = name
.replace(/\.{2,}/g, '.') // reduce multiple dots to one dot
.replace(/^\.+/, '') // remove leading dots
.replace(/\.+$/, '') // remove trailing dots
.replace(/[^\w\.]/g, '') // remove non word/dot chars
.replace(/^(.{30}).+/, '$1'); // restrict to 30 chars
console.log(name + ' => ' + clean);
});
Output:
f => f
foobar => foobar
_foo.bar => _foo.bar
_foo..bar => _foo.bar
_foo...bar => _foo.bar
_foo.bar. => _foo.bar
.foo.bar => foo.bar
foo<$^*>bar => foobar
123456789012345678901234567890 => 123456789012345678901234567890
1234567890123456789012345678901 => 123456789012345678901234567890
Note that the original regex requires at least one char. You need to test for empty string after cleanup.
I am trying to parse an array using Javascript given a string that's hyphenated.
- foo
- bar
I have gotten very close to figuring it out. I have trimmed it down to where I get the two items using this code.
const chunks = input.split(/\ ?\-\ ?/);
chunks = chunks.slice(1);
This would trim the previous input down to this.
["foo\n", "bar"]
I've tried many solutions to get the newline character out of the string regardless of the number of items in the array, but nothing worked out. It would be greatly appreciated if someone could help me solve this issue.
You could for example split, remove all the empty entries, and then trim each item to also remove all the leading and trailing whitespace characters including the newlines.
Note that you don't have to escape the space and the hyphen.
const input = `- foo
- bar`;
const chunks = input.split(/ ?- ?/)
.filter(Boolean)
.map(s => s.trim());
console.log(chunks);
Or the same approach removing only the newlines:
const input = `- foo
- bar`;
const chunks = input.split(/ ?- ?/)
.filter(Boolean)
.map(s => s.replace(/\r?\n|\r/g, ''));
console.log(chunks);
Instead of split, you might also use a match with a capture group:
^ ?- ?(.*)
The pattern matches:
^ Start of string
?- ? Match - between optional spaces
(.*) Capture group 1, match the rest of the line
const input = `- foo
- bar`;
const chunks = Array.from(input.matchAll(/^ ?- ?(.*)/gm), m => m[1]);
console.log(chunks);
You could loop over like so and remove the newline chars.
const data = ["foo\n", "bar"]
const res = data.map(str => str.replaceAll('\n', ''))
console.log(res)
Instead of trimming after the split. Split wisely and then map to replace unwanted string. No need to loop multiple times.
const str = ` - foo
- bar`;
let chunks = str.split("\n").map(s => s.replace(/^\W+/, ""));
console.log(chunks)
let chunks2 = str.split("\n").map(s => s.split(" ")[2]);
console.log(chunks2)
You could use regex match with:
Match prefix "- " but exclude from capture (?<=- ) and any number of character different of "\n" [^\n]*.
const str = `
- foo
- bar
`
console.log(str.match(/(?<=- )[^\n]*/g))
chunks.map((data) => {
data = data.replace(/(\r\n|\n|\r|\\n|\\r)/gm, "");
return data;
})
const str = ` - foo
- bar`;
const result = str.replace(/([\r\n|\n|\r])/gm, "")
console.log(result)
That should remove all kinds of line break in a string and after that you can perform other actions to get the expected result like.
const str = ` - foo
- bar`;
const result = str.replace(/([\r\n|\n|\r|^\s+])/gm, "")
console.log(result)
const actualResult = result.split('-')
actualResult.splice(0,1)
console.log(actualResult)
I trying to split text by two rules:
Split by whitespace
Split words greater than 5 symbols into two separate words like (aaaaawww into aaaaa- and www)
I create regex that can detect this rules (https://regex101.com/r/fyskB3/2) but can't understand how to make both rules work in (text.split(/REGEX/)
Currently regex - (([\s]+)|(\w{5})(?=\w))
For example initial text is hello i am markopollo and result should look like ['hello', 'i', 'am', 'marko-', 'pollo']
It would probably be easier to use .match: match up to 5 characters that aren't whitespace:
const str = 'wqerweirj ioqwejr qiwejrio jqoiwejr qwer qwer';
console.log(
str.match(/[^ ]{1,5}/g)
)
My approach would be to process the string before splitting (I'm a big fan of RegEx):
1- Search and replace all the 5 consecutive non-last characters with \1-.
The pattern (\w{5}\B) will do the trick, \w{5} will match 5 exact characters and \B will match only if the last character is not the ending character of the word.
2- Split the string by spaces.
var text = "hello123467891234 i am markopollo";
var regex = /(\w{5}\B)/g;
var processedText = text.replace(regex, "$1- ");
var result = processedText.split(" ");
console.log(result)
Hope it helps!
Something like this should work:
const str = "hello i am markopollo";
const words = str.split(/\s+/);
const CHUNK_SIZE=5;
const out = [];
for(const word of words) {
if(word.length > CHUNK_SIZE) {
let chunks = chunkSubstr(word,CHUNK_SIZE);
let last = chunks.pop();
out.push(...chunks.map(c => c + '-'),last);
} else {
out.push(word);
}
}
console.log(out);
// credit: https://stackoverflow.com/a/29202760/65387
function chunkSubstr(str, size) {
const numChunks = Math.ceil(str.length / size)
const chunks = new Array(numChunks)
for (let i = 0, o = 0; i < numChunks; ++i, o += size) {
chunks[i] = str.substr(o, size)
}
return chunks
}
i.e., first split the string into words on spaces, and then find words longer than 5 chars and 'chunk' them. I popped off the last chunk to avoid adding a - to it, but there might be a more efficient way if you patch chunkSubstr instead.
regex.split doesn't work so well because it will basically remove those items from the output. In your case, it appears you want to strip the whitespace but keep the words, so splitting on both won't work.
Uses the regex expression of #CertainPerformance = [^\s]{1,5}, then apply regex.exec, finally loop all matches to reach the goal.
Like below demo:
const str = 'wqerweirj ioqwejr qiwejrio jqoiwejr qwer qwer'
let regex1 = RegExp('[^ ]{1,5}', 'g')
function customSplit(targetString, regexExpress) {
let result = []
let matchItem = null
while ((matchItem = regexExpress.exec(targetString)) !== null) {
result.push(
matchItem[0] + (
matchItem[0].length === 5 && targetString[regexExpress.lastIndex] && targetString[regexExpress.lastIndex] !== ' '
? '-' : '')
)
}
return result
}
console.log(customSplit(str, regex1))
console.log(customSplit('hello i am markopollo', regex1))
I'm looking to reformat (replace, not validate - there are many references for validating) a phone number for display in Javascript. Here's an example of some of the data:
123 4567890
(123) 456-7890
(123)456-7890
123 456 7890
123.456.7890
(blank/null)
1234567890
Is there an easy way to use a regular expression to do this? I'm looking for the best way to do this. Is there a better way?
I want to reformat the number to the following: (123) 456-7890
Assuming you want the format "(123) 456-7890":
function formatPhoneNumber(phoneNumberString) {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(\d{3})(\d{3})(\d{4})$/);
if (match) {
return '(' + match[1] + ') ' + match[2] + '-' + match[3];
}
return null;
}
Here's a version that allows the optional +1 international code:
function formatPhoneNumber(phoneNumberString) {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(1|)?(\d{3})(\d{3})(\d{4})$/);
if (match) {
var intlCode = (match[1] ? '+1 ' : '');
return [intlCode, '(', match[2], ') ', match[3], '-', match[4]].join('');
}
return null;
}
formatPhoneNumber('+12345678900') // => "+1 (234) 567-8900"
formatPhoneNumber('2345678900') // => "(234) 567-8900"
Possible solution:
function normalize(phone) {
//normalize string and remove all unnecessary characters
phone = phone.replace(/[^\d]/g, "");
//check if number length equals to 10
if (phone.length == 10) {
//reformat and return phone number
return phone.replace(/(\d{3})(\d{3})(\d{4})/, "($1) $2-$3");
}
return null;
}
var phone = '(123)4567890';
phone = normalize(phone); //(123) 456-7890
var x = '301.474.4062';
x = x.replace(/\D+/g, '')
.replace(/(\d{3})(\d{3})(\d{4})/, '($1) $2-$3');
alert(x);
This answer borrows from maerics' answer. It differs primarily in that it accepts partially entered phone numbers and formats the parts that have been entered.
phone = value.replace(/\D/g, '');
const match = phone.match(/^(\d{1,3})(\d{0,3})(\d{0,4})$/);
if (match) {
phone = `${match[1]}${match[2] ? ' ' : ''}${match[2]}${match[3] ? '-' : ''}${match[3]}`;
}
return phone
I'm using this function to format US numbers.
function formatUsPhone(phone) {
var phoneTest = new RegExp(/^((\+1)|1)? ?\(?(\d{3})\)?[ .-]?(\d{3})[ .-]?(\d{4})( ?(ext\.? ?|x)(\d*))?$/);
phone = phone.trim();
var results = phoneTest.exec(phone);
if (results !== null && results.length > 8) {
return "(" + results[3] + ") " + results[4] + "-" + results[5] + (typeof results[8] !== "undefined" ? " x" + results[8] : "");
}
else {
return phone;
}
}
It accepts almost all imaginable ways of writing a US phone number. The result is formatted to a standard form of (987) 654-3210 x123
thinking backwards
Take the last digits only (up to 10) ignoring first "1".
function formatUSNumber(entry = '') {
const match = entry
.replace(/\D+/g, '').replace(/^1/, '')
.match(/([^\d]*\d[^\d]*){1,10}$/)[0]
const part1 = match.length > 2 ? `(${match.substring(0,3)})` : match
const part2 = match.length > 3 ? ` ${match.substring(3, 6)}` : ''
const part3 = match.length > 6 ? `-${match.substring(6, 10)}` : ''
return `${part1}${part2}${part3}`
}
example input / output as you type
formatUSNumber('+1333')
// (333)
formatUSNumber('333')
// (333)
formatUSNumber('333444')
// (333) 444
formatUSNumber('3334445555')
// (333) 444-5555
2021
libphonenumber-js
Example
import parsePhoneNumber from 'libphonenumber-js'
const phoneNumber = parsePhoneNumber('+12133734253')
phoneNumber.formatInternational() === '+1 213 373 4253'
phoneNumber.formatNational() === '(213) 373-4253'
phoneNumber.getURI() === 'tel:+12133734253'
Based on David Baucum's answer - here is a version that trys to improve auto-replacement "as you type" for example in a React onChange event handler:
function formatPhoneNumber(phoneNumber) {
const cleanNum = phoneNumber.toString().replace(/\D/g, '');
const match = cleanNum.match(/^(\d{3})(\d{0,3})(\d{0,4})$/);
if (match) {
return '(' + match[1] + ') ' + (match[2] ? match[2] + "-" : "") + match[3];
}
return cleanNum;
}
//...
onChange={e => setPhoneNum(formatPhoneNumber(e.target.value))}
It will insert (###) as soon as there are 3 numbers and then it will keep following the RegEx until it looks like this (###) ###-####
I've extended David Baucum's answer to include support for extensions up to 4 digits in length. It also includes the parentheses requested in the original question. This formatting will work as you type in the field.
phone = phone.replace(/\D/g, '');
const match = phone.match(/^(\d{1,3})(\d{0,3})(\d{0,4})(\d{0,4})$/);
if (match) {
phone = `(${match[1]}${match[2] ? ') ' : ''}${match[2]}${match[3] ? '-' : ''}${match[3]}${match[4] ? ' x' : ''}${match[4]}`;
}
return phone;
Almost all of these have issues when the user tries to backspace over the delimiters, particularly from the middle of the string.
Here's a jquery solution that handles that, and also makes sure the cursor stays in the right place as you edit:
//format text input as phone number (nnn) nnn-nnnn
$('.myPhoneField').on('input', function (e){
var $phoneField = e.target;
var cursorPosition = $phoneField.selectionStart;
var numericString = $phoneField.value.replace(/\D/g, '').substring(0, 10);
// let user backspace over the '-'
if (cursorPosition === 9 && numericString.length > 6) return;
// let user backspace over the ') '
if (cursorPosition === 5 && numericString.length > 3) return;
if (cursorPosition === 4 && numericString.length > 3) return;
var match = numericString.match(/^(\d{1,3})(\d{0,3})(\d{0,4})$/);
if (match) {
var newVal = '(' + match[1];
newVal += match[2] ? ') ' + match[2] : '';
newVal += match[3] ? '-' + match[3] : '';
// to help us put the cursor back in the right place
var delta = newVal.length - Math.min($phoneField.value.length, 14);
$phoneField.value = newVal;
$phoneField.selectionEnd = cursorPosition + delta;
} else {
$phoneField.value = '';
}
})
var numbers = "(123) 456-7890".replace(/[^\d]/g, ""); //This strips all characters that aren't digits
if (numbers.length != 10) //wrong format
//handle error
var phone = "(" + numbers.substr(0, 3) + ") " + numbers.substr(3, 3) + "-" + numbers.substr(6); //Create format with substrings
Here is one that will accept both phone numbers and phone numbers with extensions.
function phoneNumber(tel) {
var toString = String(tel),
phoneNumber = toString.replace(/[^0-9]/g, ""),
countArrayStr = phoneNumber.split(""),
numberVar = countArrayStr.length,
closeStr = countArrayStr.join("");
if (numberVar == 10) {
var phone = closeStr.replace(/(\d{3})(\d{3})(\d{4})/, "$1.$2.$3"); // Change number symbols here for numbers 10 digits in length. Just change the periods to what ever is needed.
} else if (numberVar > 10) {
var howMany = closeStr.length,
subtract = (10 - howMany),
phoneBeginning = closeStr.slice(0, subtract),
phoneExtention = closeStr.slice(subtract),
disX = "x", // Change the extension symbol here
phoneBeginningReplace = phoneBeginning.replace(/(\d{3})(\d{3})(\d{4})/, "$1.$2.$3"), // Change number symbols here for numbers greater than 10 digits in length. Just change the periods and to what ever is needed.
array = [phoneBeginningReplace, disX, phoneExtention],
afterarray = array.splice(1, 0, " "),
phone = array.join("");
} else {
var phone = "invalid number US number";
}
return phone;
}
phoneNumber("1234567891"); // Your phone number here
For all international Phone numbers with country code upto 3 digits, we can change the original answer a little bit as below.
For first match instead of looking for '1' we should look for 1-3 digits.
export const formatPhoneNumber = (phoneNumberString) => {
var cleaned = ('' + phoneNumberString).replace(/\D/g, '');
var match = cleaned.match(/^(\d{1,3}|)?(\d{3})(\d{3})(\d{4})$/);
if (match) {
var intlCode = (match[1] ? `+${match[1]} ` : '');
return [intlCode, '(', match[2], ') ', match[3], '-', match[4]].join('');
}
return null;
}
console.log( formatPhoneNumber('16464765278') )//+1 (646) 476-5278
console.log( formatPhoneNumber('+2549114765278')) //+254 (911) 476-5278
console.log( formatPhoneNumber('929876543210') )//+92 (987) 654-3210
Fulfils my requirement.
For US Phone Numbers
/^\(?(\d{3})\)?[- ]?(\d{3})[- ]?(\d{4})$/
Let’s divide this regular expression in smaller fragments to make is easy to understand.
/^\(?: Means that the phone number may begin with an optional (.
(\d{3}): After the optional ( there must be 3 numeric digits. If the phone number does not have a (, it must start with 3 digits. E.g. (308 or 308.
\)?: Means that the phone number can have an optional ) after first 3 digits.
[- ]?: Next the phone number can have an optional hyphen (-) after ) if present or after first 3 digits.
(\d{3}): Then there must be 3 more numeric digits. E.g (308)-135 or 308-135 or 308135
[- ]?: After the second set of 3 digits the phone number can have another optional hyphen (-). E.g (308)-135- or 308-135- or 308135-
(\d{4})$/: Finally, the phone number must end with four digits. E.g (308)-135-7895 or 308-135-7895 or 308135-7895 or 3081357895.
Reference :
http://www.zparacha.com/phone_number_regex/
You can use this functions to check valid phone numbers and normalize them:
let formatPhone = (dirtyNumber) => {
return dirtyNumber.replace(/\D+/g, '').replace(/(\d{3})(\d{3})(\d{4})/, '($1) $2-$3');
}
let isPhone = (phone) => {
//normalize string and remove all unnecessary characters
phone = phone.replace(/\D+/g, '');
return phone.length == 10? true : false;
}
The solutions above are superior, especially if using Java, and encountering more numbers with more than 10 digits such as the international code prefix or additional extension numbers. This solution is basic (I'm a beginner in the regex world) and designed with US Phone numbers in mind and is only useful for strings with just 10 numbers with perhaps some formatting characters, or perhaps no formatting characters at all (just 10 numbers). As such I would recomend this solution only for semi-automatic applications. I Personally prefer to store numbers as just 10 numbers without formatting characters, but also want to be able to convert or clean phone numbers to the standard format normal people and apps/phones will recognize instantly at will.
I came across this post looking for something I could use with a text cleaner app that has PCRE Regex capabilities (but no java functions). I will post this here for people who could use a simple pure Regex solution that could work in a variety of text editors, cleaners, expanders, or even some clipboard managers. I personally use Sublime and TextSoap. This solution was made for Text Soap as it lives in the menu bar and provides a drop-down menu where you can trigger text manipulation actions on what is selected by the cursor or what's in the clipboard.
My approach is essentially two substitution/search and replace regexes. Each substitution search and replace involves two regexes, one for search and one for replace.
Substitution/ Search & Replace #1
The first substitution/ search & replace strips non-numeric numbers from an otherwise 10-digit number to a 10-digit string.
First Substitution/ Search Regex: \D
This search string matches all characters that is not a digit.
First Substitution/ Replace Regex: "" (nothing, not even a space)
Leave the substitute field completely blank, no white space should exist including spaces. This will result in all matched non-digit characters being deleted. You should have gone in with 10 digits + formatting characters prior this operation and come out with 10 digits sans formatting characters.
Substitution/ Search & Replace #2
The second substitution/search and replace search part of the operation captures groups for area code $1, a capture group for the second set of three numbers $2, and the last capture group for the last set of four numbers $3. The regex for the substitute portion of the operation inserts US phone number formatting in between the captured group of digits.
Second Substitution/ Search Regex: (\d{3})(\d{3})(\d{4})
Second Substitution/ Replace Regex: \($1\) $2\-$3
The backslash \ escapes the special characters (, ) , (<-whitespace), and - since we are inserting them between our captured numbers in capture groups $1, $2, & $3 for US phone number formatting purposes.
In TextSoap I created a custom cleaner that includes the two substitution operation actions, so in practice it feels identical to executing a script. I'm sure this solution could be improved but I expect complexity to go up quite a bit. An improved version of this solution is welcomed as a learning experience if anyone wants to add to this.