I have problem with simple rexex. I have example strings like:
Something1\sth2\n649 sth\n670 sth x
Sth1\n\something2\n42 036 sth\n42 896 sth y
I want to extract these numbers from strings. So From first example I need two groups: 649 and 670. From second example: 42 036 and 42 896. Then I will remove space.
Currently I have something like this:
\d+ ?\d+
But it is not a good solution.
You can use
\n\d+(?: \d+)?
\n - Match new line
\d+ - Match digit from 0 to 9 one or more time
(?: \d+)? - Match space followed by digit one or more time. ( ? makes it optional )
let strs = ["Something1\sth2\n649 sth\n670 sth x","Sth1\n\something2\n42 036 sth\n42 896 sth y"]
let extractNumbers = str => {
return str.match(/\n\d+(?: \d+)?/g).map(m => m.replace(/\s+/g,''))
}
strs.forEach(str=> console.log(extractNumbers(str)))
If you need to remove the spaces. Then the easiest way for you to do this would be to remove the spaces and then scrape the numbers using 2 different regex.
str.replace(/\s+/, '').match(/\\n(\d+)/g)
First you remove spaces using the \s token with a + quantifier using replace.
Then you capture the numbers using \\n(\d+).
The first part of the regex helps us make sure we are not capturing numbers that are not following a new line, using \ to escape the \ from \n.
The second part (\d+) is the actual match group.
var str1 = "Something1\sth2\n649 sth\n670 sth x";
var str2 = "Sth1\n\something2\n42 036 sth\n42 896 sth y";
var reg = /(?<=\n)(\d+)(?: (\d+))?/g;
var d;
while(d = reg.exec(str1)){
console.log(d[2] ? d[1]+d[2] : d[1]);
}
console.log("****************************");
while(d = reg.exec(str2)){
console.log(d[2] ? d[1]+d[2] : d[1]);
}
Related
var cartstring = "27,00 - R"
How can I remove spaces and "-" and "R" using only regex (not allowed to use slice etc.)? I need to make strings cartstring1 and cartstring2 which should both be equal to "27,00", first by removing spaces and "-" and "R", and second by allowing only numbers and ",".
cartstring1 = cartstring.replace(/\s/g, "");
cartstring2 = cartstring.replace(/\D/g, "");
Please help me modify these regular expressions to have a working code. I tried to read about regex but still cannot quite get it. Thank you very much in advance.
you can just capture just what you are interested in number and comma:
let re = /[\d,]+/g
let result = "27,00 - R".match(re)
console.log(result)
You can group the characters you want to remove:
var cartstring = "27,00 - R"
let res = cartstring.replace(/(\s|-|R)/g, "")
console.log(res)
Or alternatively, split the string by a space and get the first item:
var cartstring = "27,00 - R"
let res = cartstring.split(" ")[0]
console.log(res)
You are using 2 replacements, one replacing all whitespace chars \s and the other replacing all non digits \D, but note that \D also matches \s so you could omit the first call.
Using \D will also remove the comma that you want to keep, so you can match all chars except digits or a comma using [^\d,]+ in a single replacement instead:
var cartstring = "27,00 - R";
console.log(cartstring.replace(/[^\d,]+/g, ''));
Example data expected output
sds-rwewr-dddd-cash0-bbb cash0
rrse-cash1-nonre cash1
loan-snk-cash2-ssdd cash2
garb-cash3-dfgfd cash3
loan-unwan-cash4-something cash4
The common pattern is here, need to extract a few chars before the last hyphen of given string.
var regex1= /.*(?=(?:-[^-]*){1}$)/g ; //output will be "ds-rwewr-dddd-cash0" from "sds-rwewr-dddd-cash0-bbb "
var regex2 = /\w[^-]*$/g ; //output will be "cash0" from "ds-rwewr-dddd-cash0"
var res =regex2.exec(regex1.exec(sds-rwewr-dddd-cash0-bbb)) //output will cash0
Although above nested regex is working as expected but may not be optimize one. So any help will be appreciated for optimized regex
You can use
/\w+(?=-[^-]*$)/
If the part before the last hyphen can contain chars other than word chars, keep using \w[^-]*: /\w[^-]*(?=-[^-]*$)/. If you do not need to check the first char of your match, simply use /[^-]+(?=-[^-]*$)/.
See the regex demo.
Details:
\w+ - one or more word chars
(?=-[^-]*$) - that must be followed with - and then zero or more chars other than - till the end of string.
JavaScript demo
const texts = ['sds-rwewr-dddd-cash0-bbb','rrse-cash1-nonre','loan-snk-cash2-ssdd','garb-cash3-dfgfd','loan-unwan-cash4-something'];
const regex = /\w+(?=-[^-]*$)/;
for (var text of texts) {
console.log(text, '=>', text.match(regex)?.[0]);
}
replacing the first and last few characters with the * character, i am able to solve the str1 case. How can i solve the remaining one. Right now i am able to mask the last 4 characters.
how can i mask the first 3 or 4 characters. ? whats wrong in the regex pattern
var str1 = "1234567890123456";
str1 = str1.replace(/\d(?=\d{4})/g, "*");
console.log(str1)
var str2 = "123-456-789-101112"
str2 = str2.replace(/\d(?=\d{4})/g, "*");
console.log(str2) // expected ***-***-***-**1112
var str3 = "abc:def:12324-12356"
str3 = str3.replace(/\d(?=\d{4})/g, "*");
console.log(str3) // expected ***:***:*****-*2356
Right now it is masking only the four characters from last, how can i mask 4 characters from front also like
1234567890123456 => 1234********3456
123-456-789-101112 => 123-4**-***-**1112
abc:def:12324-12356 => abc:d**:*****-*2356
One option is to lookahead for non-space characters followed by 4 digits. Since you want to replace the alphabetical characters too, use a character set [a-z\d] rather than just \d:
const repl = str => console.log(str.replace(/[a-z\d](?=\S*\d{4})/g, "*"));
repl("1234567890123456");
repl("123-456-789-101112");
repl("abc:def:12324-12356");
If you want to keep the first 4 alphanumeric characters as well, then it's significantly more complicated - match and capture the first 4 characters, possibly interspersed with separators, then capture the in-between characters, then capture the last 4 digits. Use a replacer function to replace all non-separator characters in the second group with *s:
const repl = str => console.log(str.replace(
/((?:[a-z\d][-#.:]?){4})([-#:.a-z\d]+)((?:[a-z\d][-#.:]?){4})/ig,
(_, g1, g2, g3) => g1 + g2.replace(/[a-z\d]/ig, '*') + g3
));
repl("1234567890123456");
repl("123-456-789-101112");
repl("abc:def:12324-12356");
repl("test#test.com");
I am trying to capture all characters between multiple instances of asterisks, which are comma delimited in a string. Here's an example of the string:
checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*
The caveat is that the phrase must start and end with an asterisk. I have been able to come close by using the following regex, however, it won't discard any matches when the captured string is missing the starting asterisk(*):
let str = "checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*"
const regex = /[^\,\*]+(?=\*)/gi;
var a = str.match(regex)
console.log(a) // answer should exclude checkboxID0 and checkboxID5
The answer returns the following, however, "checkboxID0 and checkboxID5" should be excluded as it doesn't start with an asterisk.
[
"checkboxID0",
"checkboxID1",
"checkboxID3",
"checkboxID5"
]
Thanks, in advance!
You need to use asterisks on both ends of the pattern and capture all 1 or more chars other than commas and asterisks in between:
/\*([^,*]+)\*/g
See the regex demo
Pattern details
\* - an asterisk
([^,*]+) - Capturing group 1: one or more chars other than , and *
\* - an asterisk
JS demo:
var regex = /\*([^,*]+)\*/g;
var str = "checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*";
var m, res = [];
while (m = regex.exec(str)) {
res.push(m[1]);
}
console.log(res);
How can I get the strings between last 2 slashes in regex in javascript?
for example:
stackoverflow.com/questions/ask/index.html => "ask"
http://regexr.com/foo.html?q=bar => "regexr.com"
https://www.w3schools.com/icons/default.asp => "icons"
You can use /\/([^/]+)\/[^/]*$/; [^/]*$ matches everything after the last slash, \/([^/]+)\/ matches the last two slashes, then you can capture what is in between and extract it:
var samples = ["stackoverflow.com/questions/ask/index.html",
"http://regexr.com/foo.html?q=bar",
"https://www.w3schools.com/icons/default.asp"]
console.log(
samples.map(s => s.match(/\/([^/]+)\/[^/]*$/)[1])
)
You can solve this by using split().
let a = 'stackoverflow.com/questions/ask/index.html';
let b = 'http://regexr.com/foo.html?q=bar';
let c = 'https://www.w3schools.com/icons/default.asp';
a = a.split('/')
b = b.split('/')
c = c.split('/')
indexing after split()
console.log(a[a.length-2])
console.log(b[b.length-2])
console.log(c[c.length-2])
I personally do not recommend using regex. Because it is hard to maintain
I believe that will do:
[^\/]+(?=\/[^\/]*$)
[^\/]+ This matches all chars other than /. Putting this (?=\/[^\/]*$) in the sequence looks for the pattern that comes before the last /.
var urls = [
"stackoverflow.com/questions/ask/index.html",
"http://regexr.com/foo.html?q=bar",
"https://www.w3schools.com/icons/default.asp"
];
urls.forEach(url => console.log(url.match(/[^\/]+(?=\/[^\/]*$)/)[0]));
You can use (?=[^/]*\/[^/]*$)(.*?)(?=\/[^/]*$). You can test it here: https://www.regexpal.com/
The format of the regex is: (positive lookahead for second last slash)(.*?)(positive lookahead for last slash).
The (.*?) is a lazy match for what's between the slashes.
references:
Replace second to last "/" character in URL with a '#'
RegEx that will match the last occurrence of dot in a string