Typescript Type 'number' is not assignable │ to type 'string'

Question

I am currently knee deep in making a currency formatter directive for an Angular 4 app. on the parse strip out everything other than the numbers and the decimal and end up with a stringified float, but I need it to return as a float so I can do math with it.

parse(value: string, fractionSize: number = 2): number {
  let val = value.replace(/([^0-9.])+/ig, '');
  let [ integer, fraction = "" ] = (val || "").split(this.DECIMAL_SEPARATOR);
  integer = integer.replace(new RegExp(this.THOUSANDS_SEPARATOR, "g"), "");
  fraction = parseInt(fraction, 10) > 0 && fractionSize > 0
    ? this.DECIMAL_SEPARATOR + (fraction + PADDING).substring(0, fractionSize)
    : "";
  let result = `${integer}${fraction}`;
  // at this point result = "100.55";
  result = parseFloat(result); // this refuses to compile saying "Type 'number' is not assignable │ to type 'string'"
  return result;
}

CRice · Accepted Answer · 2018-02-07 17:45:14Z

10

The two lines:

let result = `${integer}${fraction}`;
result = parseFloat(result);

are the problem. Typescript is pretty good about infering the type of a variable when it's not explicitly declared. In this case, because you assign a string to the result, typescript infers it's type as a string. To fix this, you have two options. First, explicitly declare the type of that variable so that it allows both strings and numbers:

let result: string|number = `${integer}${fraction}`;
result = parseFloat(result); // now should be ok.

Or you can assign the parsed number to a new variable, instead of reusing the result variable:

let result = `${integer}${fraction}`;
let numberResult = parseFloat(result); // now should be ok.

answered Feb 7, 2018 at 17:45

CRice

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Comments

Wenfang Du · Accepted Answer · 2021-06-12 23:25:07Z

3

You can't assign different types to a variable in Typescript. If you initialized the variable with a string it must remain a string.

    let result = `${integer}${fraction}`;
    let resultAsNumber = parseFloat(result); 
    return resultAsNumber

This is a common cause of errors and the type system tries to prevent you from doing this.

edited Jun 12, 2021 at 23:25

Wenfang Du

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answered Feb 7, 2018 at 17:44

Titian Cernicova-Dragomir

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Comments

Sajeetharan · Accepted Answer · 2018-02-07 18:05:41Z

0

There is an obvious error, you are trying to assign a number to a string variable, no need to parse to Float again, just return the number by parsing it

let result = `${integer}${fraction}`;
  // at this point result = "100.55";
return parseFloat(result);

edited Feb 7, 2018 at 18:05

answered Feb 7, 2018 at 17:41

Sajeetharan

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