Replace specific group of characters in match using regex

Question

So I want to match in string bellow, which will be formula for excell/ spreadsheet type cell, all addresses: \w+\d+ , and change numbers only number part in them. I want to get following strings from original: "= A2 + B2", "=A3+B3", "=A4+B4" ...

I tried:

const a = "=A1 + B1"
for (let i = 0; i < 100 ; i++) {
    const b = a.replace(/\w+$(\d+)/g, String(i + 1));
    console.log(b)
}

and it gives result:

then if I do without $ before grouping () parentesis:

const a = "=A1 + B1"
for (let i = 0; i < 100 ; i++) {
    const b = a.replace(/\w+(\d+)/g, String(i + 1));
    console.log(b)
}

I get:

So I want number (which in string is 1, but can be anything, like 24, 67, ) to be changed with index i + 1 in for loop. — Paul Gutsal
– Paul Gutsal, Commented Jan 2, 2023 at 13:18
yes, But the idea is that user insert string =A1 + B1, and it will be adapted for differant cell adresses with this regex. — Paul Gutsal
– Paul Gutsal, Commented Jan 2, 2023 at 13:25

Joan Lara · Accepted Answer · 2023-01-02 13:21:15Z

1

Something like this?

const a = "=A1 + B1"
for (let i = 0; i < 10 ; i++) {
    const b = a.replace(/\b\w+\d+\b/g, function(match) {
        const num = match.match(/\d+/);
        const newNum = Number(num[0]) + i;
        return match.replace(/\d+/, newNum);
    });
    console.log(b);
}

answered Jan 2, 2023 at 13:21

Joan Lara

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Comments

T.J. Crowder · Accepted Answer · 2023-01-02 13:28:10Z

From your comment, it sounds like you want 100 results, replacing the row numbers with 1, 2, 3, etc.

If so, capture the column number and use it in the replacement:

function updateFormula(formula, i) {
    const result = formula.replace(
        /\b([A-Z]+)\d+\b/g,
        (m, column) => column + i
    );
    return result;
}

const formula = "=A42 + B27";
for (let n = 1; n <= 100; ++n) {
    const result = updateFormula(formula, n);
    // ...
}

Live Example (only doing 20 because the in-snippet console drops rows after a while):

function updateFormula(formula, i) {
    const result = formula.replace(
        /\b([A-Z]+)\d+\b/g,
        (m, column) => column + i
    );
    return result;
}

const formula = "=A42 + B27";
for (let n = 1; n <= 20; ++n) {
    const result = updateFormula(formula, n);
    console.log(`${JSON.stringify(formula)} => ${JSON.stringify(result)}`);
}

.as-console-log {
    max-height: 100% !important;
}

Paul Gutsal · Accepted Answer · 2023-01-02 13:48:49Z

0

So, all answers are really great, but in code I decided to implement something similar to Joan Lara Ganau answere.

const a = "=A1 + B1"
for (let i = 0; i < 100 ; i++) {
    const num = String(i+1);
    const b = a.replace(/\w+\d+/g, function(match) {
        return match.replace(/\d+/g, num)
    });
    console.log(b);
}

which produces :

Still don't know why we need \b, is it similar to ()?

answered Jan 2, 2023 at 13:48

Paul Gutsal

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Comments

Thomas · Accepted Answer · 2023-01-02 13:59:44Z

0

Just match the digits after a letter.

const a = "=A1 + B1";
for (let i = 0; i < 10; i++) {
  const b = a.replace(/(?<=[A-Z]+)\d+/g, i + 1);
  console.log(b);
}

Or you can use split the string to get the pattern, so you don't have to match it on every iteration.

const a = "=A1 + B1"
const parts = a.split(/(?<=[A-Z]+)\d+/g);
for (let i = 0; i < 10; i++) {
  const b = parts.join(i + 1);
  console.log(b);
}

answered Jan 2, 2023 at 13:59

Thomas

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