Javascript 将变量视为字符串，为什么？

Question

I have the variable y, which is a subtotal. Its value is different depending on what happens with the html, but throughout the script I declared it like this:

var y = 21.78;

etc. Why is it that on my last equation where I add up the total, it treats them as strings when I want to add the values?

var tax = (0.055*y).toFixed(2);
var totalprice = y+tax;
/* totalprice holds "21.781.20" instead of 22.98 */

Answer 1

According to:

https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Number/toFixed

toFixed() returns:

A string representation of number that does not use exponential notation and has exactly digits digits after the decimal place.

thus, y+tax is cast to a string since one of the operands is a string.

In my opinion, this would make sense as Javascript's intrinsic numeric types do not have the ability to store a specific number of decimal place digits, so a string would be the most appropriate data structure to store this with.

I would advise you do all your addition before calling toFixed() , since the method is most suitable for formatting display output.

 var taxRate = 0.055; var subtotal = 21.78; var tax = (taxRate * subtotal).toFixed(2), totalprice = ((1+taxRate) * subtotal).toFixed(2); document.write(totalprice);

Answer 2

The.toFixed() method returns a string. Try applying that method as the last step after all other calculations.

Answer 3

Here's a simple fix. Put '+' in front of the tax variable to convert it to a number.

var y = 21.78;
var tax = (0.055*y).toFixed(2);
var totalprice = y+ (+tax);
totalprice === 22.98;

If you don't want any rounding errors when you use toFixed, then include this re-implementation of it in your script. http://bateru.com/news/2012/03/reimplementation-of-number-prototype-tofixed/

Answer 4

In my experience, if there's any chance available, Javascript will see the "+" sign as concatenate rather than addition. It's driven me nuts on more than one occasion. I will generally do this rather than chance concatenation:

var totalprice = parseInt(y)+parseInt(tax);

Answer 5

When letter replaces value, multiply with 1 when you're in need of + .

var totalprice = (y*1) + tax . 

Other operands work fine, it's just the + operand that needs special treatment when variable replace value.