Check Whether an Input String Contains a Number in JavaScript

How can I check if a string is a valid number?

2nd October 2020: note that many bare-bones approaches are fraught with subtle bugs (eg. whitespace, implicit partial parsing, radix, coercion of arrays etc.) that many of the answers here fail to take into account. The following implementation might work for you, but note that it does not cater for number separators other than the decimal point ".":

function isNumeric(str) {
  if (typeof str != "string") return false // we only process strings!  
  return !isNaN(str) && // use type coercion to parse the _entirety_ of the string (`parseFloat` alone does not do this)...
         !isNaN(parseFloat(str)) // ...and ensure strings of whitespace fail
}

This works regardless of whether the variable content is a string or number.

isNaN(num)         // returns true if the variable does NOT contain a valid number

Examples

isNaN(123)         // false
isNaN('123')       // false
isNaN('1e10000')   // false (This translates to Infinity, which is a number)
isNaN('foo')       // true
isNaN('10px')      // true
isNaN('')          // false
isNaN(' ')         // false
isNaN(false)       // false

Of course, you can negate this if you need to. For example, to implement the IsNumeric example you gave:

function isNumeric(num){
  return !isNaN(num)
}

Only works if the string only contains numeric characters, else it returns NaN.

+num               // returns the numeric value of the string, or NaN 
                   // if the string isn't purely numeric characters

Examples

+'12'              // 12
+'12.'             // 12
+'12..'            // NaN
+'.12'             // 0.12
+'..12'            // NaN
+'foo'             // NaN
+'12px'            // NaN

Useful for converting '12px' to 12, for example:

parseInt(num)      // extracts a numeric value from the 
                   // start of the string, or NaN.

Examples

parseInt('12')     // 12
parseInt('aaa')    // NaN
parseInt('12px')   // 12
parseInt('foo2')   // NaN      These last three may
parseInt('12a5')   // 12       be different from what
parseInt('0x10')   // 16       you expected to see.

Bear in mind that, unlike +num, parseInt (as the name suggests) will convert a float into an integer by chopping off everything following the decimal point (if you want to use parseInt() because of this behaviour, you're probably better off using another method instead):

+'12.345'          // 12.345
parseInt(12.345)   // 12
parseInt('12.345') // 12

Empty strings may be a little counter-intuitive. +num converts empty strings or strings with spaces to zero, and isNaN() assumes the same:

+''                // 0
+'   '             // 0
isNaN('')          // false
isNaN('   ')       // false

But parseInt() does not agree:

parseInt('')       // NaN
parseInt('   ')    // NaN

Check whether an input string contains a (signed) number

You can easily use a regex expression to match the numbers in the string:

function f2(input) {      let str = String(input);      let result = str.match(/\-?\d+\.\d+|\-?\d+/g)      return result    }        let result = f2("a1 12 13.b -14.5+2");        console.log(result);

let isnum = /^\d+$/.test(val);

typeof "Hello World"; // string
typeof 123;           // number

var toString = Object.prototype.toString;

_.isString = function (obj) {
  return toString.call(obj) == '[object String]';
}

_.isString("Jonathan"); // true
_.isString(new String("Jonathan")); // true

let myString = "/example/123/anotherstring" let newString = myString.replace(/\d+/g, "replaced")console.log(newString)
 myString = "/example/anotherstring/123" newString = myString.replace(/\d+/g, "replaced") console.log(newString)

function hasNumber( str ){
  return /\d/.test( str );
}

hasNumber(0)     // true
hasNumber("0")   // true
hasNumber(2)     // true
hasNumber("aba") // false
hasNumber("a2a") // true

function hasNumber( str ){
  return str.match( /\d+/g );
}

hasNumber( "11a2 b3 c" );  // ["11", "2", "3"]
hasNumber( "abc" );        // null
if( hasNumber( "a2" ) )    // "truthy statement"