Javascript 如何检查数字是浮点数还是整数？

How do I check that a number is float or integer?

提问by coure2011

How to find that a number is floator integer?

如何找到一个数字是float或integer？

1.25 --> float  
1 --> integer  
0 --> integer  
0.25 --> float

回答by kennebec

check for a remainder when dividing by 1:

除以 1 时检查余数：

function isInt(n) {
   return n % 1 === 0;
}

If you don't know that the argument is a number you need two tests:

如果你不知道参数是一个数字，你需要两个测试：

function isInt(n){
    return Number(n) === n && n % 1 === 0;
}

function isFloat(n){
    return Number(n) === n && n % 1 !== 0;
}

Update 20195 years after this answer was written, a solution was standardized in ECMA Script 2015. That solution is covered in this answer.

在编写此答案 5 年后更新 2019，在 ECMA Script 2015 中标准化了一个解决方案。此答案涵盖了该解决方案。

回答by Dagg Nabbit

Try these functions to test whether a value is a number primitive value that has no fractional part and is within the size limits of what can be represented as an exact integer.

试试这些函数来测试一个值是否是一个没有小数部分的数字原始值，并且在可以表示为一个精确整数的大小限制内。

function isFloat(n) {
    return n === +n && n !== (n|0);
}

function isInteger(n) {
    return n === +n && n === (n|0);
}

回答by warfares

Why not something like this:

为什么不是这样的：

var isInt = function(n) { return parseInt(n) === n };

回答by paperstreet7

There is a method called Number.isInteger()which is currently implemented in everything but IE. MDNalso provides a polyfill for other browsers:

有一个被调用的方法Number.isInteger()，它目前在除 IE 之外的所有东西中都实现了。MDN还为其他浏览器提供了 polyfill：

Number.isInteger = Number.isInteger || function(value) {
  return typeof value === 'number' && 
    isFinite(value) && 
    Math.floor(value) === value;
};

However, for most uses cases, you are better off using Number.isSafeIntegerwhich also checks if the value is so high/low that any decimal places would have been lost anyway. MDNhas a polyfil for this as well. (You also need the isIntegerpollyfill above.)

但是，对于大多数用例，最好使用Number.isSafeIntegerwhich 还检查值是否太高/太低以至于任何小数位都会丢失。MDN也为此提供了一个 polyfil。（您还需要isInteger上面的pollyfill。）

if (!Number.MAX_SAFE_INTEGER) {
    Number.MAX_SAFE_INTEGER = 9007199254740991; // Math.pow(2, 53) - 1;
}
Number.isSafeInteger = Number.isSafeInteger || function (value) {
   return Number.isInteger(value) && Math.abs(value) <= Number.MAX_SAFE_INTEGER;
};

回答by Marcio Mazzucato

You can use a simple regular expression:

您可以使用一个简单的正则表达式：

function isInt(value) {

    var er = /^-?[0-9]+$/;

    return er.test(value);
}

Or you can use the below functions too, according your needs. They are developed by the PHPJS Project.

或者您也可以根据需要使用以下功能。它们由PHPJS 项目开发。

is_int()=> Check if variable type is integer and if its content is integer

is_int()=> 检查变量类型是否为整数以及其内容是否为整数

is_float()=> Check if variable type is float and if its content is float

is_float()=> 检查变量类型是否为浮点型以及其内容是否为浮点型

ctype_digit()=> Check if variable type is string and if its content has only decimal digits

ctype_digit()=> 检查变量类型是否为字符串以及其内容是否只有十进制数字

Update 1

更新 1

Now it checks negative numbers too, thanks for @ChrisBartley comment!

现在它也检查负数，感谢@ChrisBartley 评论！

回答by Tal Liron

Here are efficient functions that check if the value is a number or can be safely converted toa number:

以下是检查值是否为数字或可以安全地转换为数字的高效函数：

function isNumber(value) {
    if ((undefined === value) || (null === value)) {
        return false;
    }
    if (typeof value == 'number') {
        return true;
    }
    return !isNaN(value - 0);
}

And for integers (would return false if the value is a float):

对于整数（如果值为浮点数，则返回 false）：

function isInteger(value) {
    if ((undefined === value) || (null === value)) {
        return false;
    }
    return value % 1 == 0;
}

The efficiency here is that parseInt (or parseNumber) are avoided when the value already is a number. Both parsing functions alwaysconvert to string first and then attempt to parse that string, which would be a waste if the value already is a number.

这里的效率是当值已经是数字时避免使用 parseInt（或 parseNumber）。两个解析函数总是先转换为字符串，然后尝试解析该字符串，如果值已经是数字，这将是一种浪费。

Thank you to the other posts here for providing further ideas for optimization!

感谢这里的其他帖子为优化提供了进一步的想法！

回答by shime

function isInteger(x) { return typeof x === "number" && isFinite(x) && Math.floor(x) === x; }
function isFloat(x) { return !!(x % 1); }

// give it a spin

isInteger(1.0);        // true
isFloat(1.0);          // false
isFloat(1.2);          // true
isInteger(1.2);        // false
isFloat(1);            // false
isInteger(1);          // true    
isFloat(2e+2);         // false
isInteger(2e+2);       // true
isFloat('1');          // false
isInteger('1');        // false
isFloat(NaN);          // false
isInteger(NaN);        // false
isFloat(null);         // false
isInteger(null);       // false
isFloat(undefined);    // false
isInteger(undefined);  // false

回答by Deepak Yadav

function isInt(n) 
{
    return n != "" && !isNaN(n) && Math.round(n) == n;
}
function isFloat(n){
    return n != "" && !isNaN(n) && Math.round(n) != n;
}

works for all cases.

适用于所有情况。

回答by Claudiu

As others mentioned, you only have doubles in JS. So how do you define a number being an integer? Just check if the rounded number is equal to itself:

正如其他人提到的，您在 JS 中只有双打。那么如何定义一个数是整数呢？只需检查四舍五入的数字是否等于自身：

function isInteger(f) {
    return typeof(f)==="number" && Math.round(f) == f;
}
function isFloat(f) { return typeof(f)==="number" && !isInteger(f); }

回答by Goehybrid

How about this one?

这个怎么样？

isFloat(num) {
    return typeof num === "number" && !Number.isInteger(num);
}