validate regexp based on the variable after decimal point

Question

var precision = !!values.amountprecision && values.amountprecision || '2';
var temp = /^\d+(\.\d{1,2})?$/;
temp.test(1.1221);

In the above expression, i have explicitly mentioned 1 to 2 values after decimal points. but i want to set the digits after decimal point should be based on the precision variable.

Testing numbers coerced to strings using regular expressions makes me go ick. What would you say is the precision of [`0.1 + 0.2`](https://stackoverflow.com/questions/588004/is-floating-point-math-broken)? At least round them to like five or six digits so you lose the artifacts near epsilon. Also, if you demand precision 2, `1.7` will always fail (since `1.70` will auto-coerce to `"1.7"`, not `"1.70"`). I answered the question about how to embed dynamic things into regexp; but unless you're doing something weird and wonderful that I didn't expect, you're probably want to rethink your approach. — Amadan, Oct 16 '18 at 10:31

score 1 · Accepted Answer · answered Oct 16 '18 at 10:28

1

Use the excplicit constructor instead of the literal:

var temp = new RegExp('^\d+(\.\d{' + precision + '})?$');

answered Oct 16 '18 at 10:28

Amadan

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score 0 · Answer 2 · answered Oct 16 '18 at 16:38

According your code 「var temp = /^\d+(.\d{1,2})?$/;」, I guess you need an integer or a float with 1 or 2 digits after the decimal point. And the number of digits determined by a variable, precision. If so, try this

let temp = new RegExp(`^\\d+(\\.\\d{${precision}})?$`);

validate regexp based on the variable after decimal point

2 Answers2