我如何检查如果一个变量是一个整数在JavaScript,并抛出一个警告,如果它不是?我试过了,但不管用:

<html>
    <head>
        <script type="text/javascript">
            var data = 22;
            alert(NaN(data));
        </script>
    </head>
</html>

当前回答

var x = 1.5;
if(!isNaN(x)){
 console.log('Number');
 if(x % 1 == 0){
   console.log('Integer');
 }
}else {
 console.log('not a number');
}

其他回答

你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);
}
function isInteger(argument) { return argument == ~~argument; }

用法:

isInteger(1);     // true<br>
isInteger(0.1);   // false<br>
isInteger("1");   // true<br>
isInteger("0.1"); // false<br>

or:

function isInteger(argument) { return argument == argument + 0 && argument == ~~argument; }

用法:

isInteger(1);     // true<br>
isInteger(0.1);   // false<br>
isInteger("1");   // false<br>
isInteger("0.1"); // false<br>

你也可以试试这种方法

var data = 22;
if (Number.isInteger(data)) {
    console.log("integer");
 }else{
     console.log("not an integer");
 }

or

if (data === parseInt(data, 10)){
    console.log("integer");
}else{
    console.log("not an integer");
}

接受的答案不为我工作,因为我需要检查int/float和字母。因此,尝试这将工作的int/float和字母检查

function is_int(value){
        if( (parseInt(value) % 1 === 0 )){
            return true;
        }else{
            return false;
        }
}

使用

is_int(44);   // true
is_int("44");   // true
is_int(44.55);  // true
is_int("44.55");  // true
is_int("aaa");  // false  

ecmascript -6之前最简单、最清晰的解决方案(它也足够健壮,即使传递给函数的是一个非数字值,如字符串或null,也会返回false)如下:

function isInteger(x) { return (x^0) === x; } 

下面的解决方案也可以工作,尽管没有上面的那样优雅:

function isInteger(x) { return Math.round(x) === x; }

注意,在上述实现中Math.ceil()或Math.floor()也可以同样好地使用(而不是Math.round())。

或者:

function isInteger(x) { return (typeof x === 'number') && (x % 1 === 0); }

一个相当常见的错误解决方案是:

function isInteger(x) { return parseInt(x, 10) === x; }

While this parseInt-based approach will work well for many values of x, once x becomes quite large, it will fail to work properly. The problem is that parseInt() coerces its first parameter to a string before parsing digits. Therefore, once the number becomes sufficiently large, its string representation will be presented in exponential form (e.g., 1e+21). Accordingly, parseInt() will then try to parse 1e+21, but will stop parsing when it reaches the e character and will therefore return a value of 1. Observe:

> String(1000000000000000000000)
'1e+21'

> parseInt(1000000000000000000000, 10)
1

> parseInt(1000000000000000000000, 10) === 1000000000000000000000
false