我希望在与旧的VB6 IsNumeric()函数相同的概念空间中有什么东西?
当前回答
有些人也可能从基于正则表达式的答案中受益。这里是:
一行是Integer:
const isInteger = num => /^-?[0-9]+$/.test(num+'');
一行是Numeric:接受整数和小数
const isNumeric = num => /^-?[0-9]+(?:\.[0-9]+)?$/.test(num+'');
其他回答
我的尝试有点混乱,Pherhaps不是最好的解决方案
function isInt(a){
return a === ""+~~a
}
console.log(isInt('abcd')); // false
console.log(isInt('123a')); // false
console.log(isInt('1')); // true
console.log(isInt('0')); // true
console.log(isInt('-0')); // false
console.log(isInt('01')); // false
console.log(isInt('10')); // true
console.log(isInt('-1234567890')); // true
console.log(isInt(1234)); // false
console.log(isInt('123.4')); // false
console.log(isInt('')); // false
// other types then string returns false
console.log(isInt(5)); // false
console.log(isInt(undefined)); // false
console.log(isInt(null)); // false
console.log(isInt('0x1')); // false
console.log(isInt(Infinity)); // false
2019:包括ES3、ES6和TypeScript示例
也许这已经被重复了太多次了,但是我今天也和这一个进行了斗争,并想发布我的答案,因为我没有看到任何其他答案能如此简单或彻底地做到这一点:
ES3
var isNumeric = function(num){
return (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
}
ES6
const isNumeric = (num) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
字体
const isNumeric = (num: any) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num as number);
这似乎很简单,涵盖了我在许多其他帖子中看到的所有基础,并自己思考:
// Positive Cases
console.log(0, isNumeric(0) === true);
console.log(1, isNumeric(1) === true);
console.log(1234567890, isNumeric(1234567890) === true);
console.log('1234567890', isNumeric('1234567890') === true);
console.log('0', isNumeric('0') === true);
console.log('1', isNumeric('1') === true);
console.log('1.1', isNumeric('1.1') === true);
console.log('-1', isNumeric('-1') === true);
console.log('-1.2354', isNumeric('-1.2354') === true);
console.log('-1234567890', isNumeric('-1234567890') === true);
console.log(-1, isNumeric(-1) === true);
console.log(-32.1, isNumeric(-32.1) === true);
console.log('0x1', isNumeric('0x1') === true); // Valid number in hex
// Negative Cases
console.log(true, isNumeric(true) === false);
console.log(false, isNumeric(false) === false);
console.log('1..1', isNumeric('1..1') === false);
console.log('1,1', isNumeric('1,1') === false);
console.log('-32.1.12', isNumeric('-32.1.12') === false);
console.log('[blank]', isNumeric('') === false);
console.log('[spaces]', isNumeric(' ') === false);
console.log('null', isNumeric(null) === false);
console.log('undefined', isNumeric(undefined) === false);
console.log([], isNumeric([]) === false);
console.log('NaN', isNumeric(NaN) === false);
您还可以尝试自己的isNumeric函数,并在这些用例中刚刚过去,然后扫描所有用例的“true”。
或者,查看每个返回的值:
这样对我来说很有用。
function isNumeric(num){
let value1 = num.toString();
let value2 = parseFloat(num).toString();
return (value1 === value2);
}
console.log(
isNumeric(123), //true
isNumeric(-123), //true
isNumeric('123'), //true
isNumeric('-123'), //true
isNumeric(12.2), //true
isNumeric(-12.2), //true
isNumeric('12.2'), //true
isNumeric('-12.2'), //true
isNumeric('a123'), //false
isNumeric('123a'), //false
isNumeric(' 123'), //false
isNumeric('123 '), //false
isNumeric('a12.2'), //false
isNumeric('12.2a'), //false
isNumeric(' 12.2'), //false
isNumeric('12.2 '), //false
)
报价:
isNaN(num)//如果变量不包含有效数字,则返回true
如果您需要检查前导/尾随空格,则不完全正确-例如,当需要一定数量的数字时,并且您需要获得例如“1111”而不是“111”或“111”以获得PIN输入。
更好使用:
var num = /^\d+$/.test(num)
以下内容如何?
const a = '1'
const isNumber = (a) => Number(a) === +a