How do I use regex to replace periods in a string?

Question

I'm using javascript.

I have a string:

let x = ".175\" x 2.5\" x 144\""; // .175" x 2.5" x 144"

I would like to manipulate the string to return 3 separate variables

var thickness = 0.175
var width = 2.5
var length = 144

I need to change the thickness in the string from.175 to 0.175, here is my attempt:

     let x = ".175\" x 2.5\" x 144\"";
    let regex = / /g;
    let regex2 = /\./i;
    let regex3 = /\"/g;
    //Do the regex function

    const regexFun = () => {
        try {
            console.log(x);
            if (x.charAt(0) == "."){
                const fun1 = x.replace(regex2, '0.'); 
                console.log(fun1);
            }
            else{

                const fun1 = x.replace(regex3, '');
                console.log(fun1);
            }

        } catch (err) {
            console.log(err.message)
        }
    }

In the example I gave I can delete the quotes using " but it doesn't seem to work with. How can I go about this? Thanks!

Answer 1

You could match the format of the string with named capture groups.

^(?<thickness>\d*\.?\d+)"\s+x\s+(?<width>\d*\.?\d+)"\s+x\s+(?<length>\d*\.?\d+)"$

The pattern matches:

• ^ Start of string
• (?<thickness>\d*\.?\d+) Group thickness match optional digits, optional dot and 1+ digits
• "\s+x\s+ Match " and an x char between optional whitespace chars
• (?<width>\d*\.?\d+) Group width with the same digits pattern
• "\s+x\s+ Match the " and x char
• (?<length>\d*\.?\d+) Group length with the same digits pattern
• " Match literally
• $ End of string

See the group values on a regex101 demo .

 let x = ".175\" x 2.5\" x 144\""; const regex = /^(?<thickness>\d*\.?\d+)"\s+x\s+(?<width>\d*\.?\d+)"\s+x\s+(?<length>\d*\.?\d+)"$/; const m = x.match(regex); if (m) { const thickness = parseFloat(m.groups.thickness); const width = parseFloat(m.groups.width); const length = parseFloat(m.groups.length); console.log(thickness, width, length); }

Answer 2

Use this Regex

(^|\D)(\.\d+)

(         // Begin capture
    ^     // Beginning of line
    |     // Or
    \D    // Any non-digit
)         // End capture
(         // Begin capture
    \.    // Decimal point
    \d+   // Number
)         // End capture

Replace with $10$2

$1        // Contents of the 1st capture
0         // Literal 0
$2        // Contents of the 2nd capture

 let x = ".175\" x 2.5\" x 144\""; console.log(x.replace(/(^|\D)(\.\d+)/g, "$10$2"));

Test Here

Answer 3

As Felix commented, best way may be to extract your numbers using regex first and then parse them as floats:

 var x = ".175\" x 2.5\" x 144\""; //.175" x 2.5" x 144" let numbers = x.match(/(\d*\.)?\d+/g).map(n => parseFloat(n)) var thickness = numbers[0] var width = numbers[1] var length = numbers[2] console.log('thickness: ' + thickness) console.log('width: ' + width) console.log('length: ' + length)

Answer 4

try using split() and parseFloat

 let x = ".175\" x 2.5\" x 144\""; const regexFun = () => { numbers = x.split('x').map(n=>parseFloat(n)) console.log(numbers) } regexFun()