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ViewsJS Negative lookbehind in IE11
Before asking this question, I looked into this, Previous Question
Currently, I need to find a regex that extracts any digits that are not preceded by $ symbol and it needs to work in IE which is why negative lookbehind can't be utilized.
My test strings is following
2 burgers and 3 drinks would be $15
The cost of 2 burgers and 3 drinks are $15
I want to retrieve 2,3 from the above two strings.
I have developed my regex as following by reading the previous post as following
/((?!([\$]))^)\d+|(?:[^\$])\b\d+/gm
I am not entirely sure if this is the correct one. If it is the correct one, the second issue is it is picking up the spaces. Even though I can replace the spaces later but I was wondering if there is a way to pick up the digits without spaces
var x = 'The cost of 2 burgers and 3 drinks are $15'
var y = x.match(/((?!([\$]))^)\d+|(?:[^\$])\b\d+/gm);
var z = y.toString().replace(/ /gm,'');
Thank you in advance.
3 answers
Answer question0
You can extract the numbers by matching prices and matching and capturing other numbers while only collecting the latter:
var x = 'The cost of 2 burgers and 3 drinks are $15'
var re = /\$\d+(?:\.\d+)?|(\d+(?:\.\d+)?)/g
var output=[], m;
while (m =re.exec(x)) {
if (m[1]) {
output.push(m[1])
}
}
console.log(output); // => [ "2", "3" ]See the regex demo. Here,
\$\d+(?:\.\d+)?- matches$, one or more digits, and then an optional occurrence of.and one or more digits|- or(\d+(?:\.\d+)?)- Capturing group 1: one or more digits, and then an optional occurrence of.and one or more digits.
0
You could just use some standard js array / string manipulation to get the desired output:
const strings = [
'2 burgers and 3 drinks would be $15',
'The cost of 2 burgers and 3 drinks are $15',
'The foo did bar to baz'
];
const results = strings.map(str =>
(str.match(/.?\d+/g) || []) // Match all numbers with their preceding character
.filter(s => s[0] !== '$') // Filter out matches that start with '$'
.map(s => s.trim())); // Trim the results.
console.log(results);One massive advantage here is readability. If you come back to this code a year from now, it's pretty trivial to figure out what it does.
0
If the digits should be preceded by a space, you can match either a space or assert the start of the string using (?:\s|^) and capture the digits in group 1 using (\d+).
Then you can extract the digits using group 1, which is denoted by m[1] in the example code.
(?:\s|^)(\d+)
const regex = /(?:\s|^)(\d+)/g;
[
'2 burgers and 3 drinks would be $15',
'The cost of 2 burgers and 3 drinks are $15'
].forEach(s => console.log(Array.from(s.matchAll(regex), m => m[1])))If there can be an other character before it that is not a dollar sign or a digit to prevent a partial match.
(?:[^\d$]|^)(\d+)
const regex = /(?:[^\d$]|^)(\d+)/g;
[
'2 burgers and 3 drinks would be $15',
'The cost of 2 burgers and 3 drinks are $15 A12'
].forEach(s => console.log(Array.from(s.matchAll(regex), m => m[1])))