JS: Filter array only for non-empty and type of string values

Question

I am trying to filter an array like this:

array.filter(e => { return e })

With this I want to filter all empty strings including undef

Answer 1

You can check the type of the elements using typeof:

array.filter(e => typeof e === 'string' && e !== '')

Since '' is falsy, you could simplify by just testing if e was truthy, though the above is more explicit

array.filter(e => typeof e === 'string' && e)

const array = [null, undefined, '', 'hello', '', 'world', 7, ['some', 'array'], null]

console.log(
  array.filter(e => typeof e === 'string' && e !== '')
)

Answer 2

You could check for a string and empty both in your filter method:

array.filter(e => (typeof e === 'string') && !!e)

Note: !!e returns false if the element is null, undefined, '' or 0.

I should mention that the "arrow"-function syntax only works in browsers that support ES6 or higher.

The alternative is:

array.filter(function(e) {
    return (typeof e === 'string') && !!e;
});

Note: Keep in mind that Array.prototype.filter doesn't exist in older browsers.

Answer 3

When returning a method that consists of one line as a callback in es6 there is no need for return as this happens implicitly.

Hope this helps :-)

let arr = ["", "", "fdsff", [], null, {}, undefined];

let filteredArr = arr.filter(item => (typeof item === "string" && !item) || !item)
                  
console.log(filteredArr)

Answer 4

const justStrings = array.filter(element => 
    (typeof element === 'string' || element instanceof String)
    && element
)

Explanation

To be shure your element is a string you have to check that it isn't a variable type (let str = 'hello') or an instance of String (new String('hello')) because this case would return object by typeof element.

Additionally you have to check if your element exists.