![]() But remember that toFixed() changes the number back to string unlike solution. With toFixed(), you just change the argument. This may become a problem if you want to change decimal places in future, you will have to change multiplication/division factor for every variable. Var sum = (a+b).toFixed(2) //change argument in toFixed() as you needĪnother alternative to this was given by which will need you to multiply and then later divide every value by 100. If you need to get 2 decimal places always, append another function toFixed(decimal places). Also, it will it will strip off any extra zeroes at the end after decimal point(ex: 17500.50 becomes 17500.0.00 becomes 17500). Note that a radix is not required in parseFloat(), it will always return a decimal/base10 value. This function will round off the string to nearest integer value. It works without a radix but it is always a good idea to specify this because you never know how browsers may behave(for example, see comment Namir). For this you can use parseInt(string, radix). If I go by the title of your question, you need an integer. This should give you correct result even if your number is fractional like 1,700.55. Use regex inside replace() to get rid of each appearance of. From the variables in OP, it appears that there may be fractional numbers too, so, parseInt() may not work well in such cases(digits after decimal will be stripped off). ![]() Try using replace() to replace a, with nothing and then parseFloat() to get the number as float. He pointed out that parseFloat also does not take a radix so it is always in base10. A good example of this was written by below. ![]() I assumed you did as the title was convert to an integer. ParseFloat could also be used if you do not want rounding. So, if you pass anything other than the permutation of the characters '0123456789.E-', you will get an exception. Here is my favorite: 2 places: `num = parseInt(num * 100) / 100 `ģ places: `num = parseInt(num * 1000) / 1000 `įor more information on parseInt look at mdn. 1 try to log an error in catch block to see if really nothing is happening Mohamed Ali RACHID at 0:26 1 Integer.parseInt () expects a string with only numeric characters. If you need decimal places there are a couple of tricks you can use. ParseInt always rounds to the nearest integer.
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