JS：仅针对非空和字符串值类型过滤数组

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 undefined and null . Unfortunately my array have some arrays, which should not be there. So I need also to check only for string values and remove all other.

How do I do that?

Answer 1

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 2

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 3

>[1,,3,,5].filter(String)
[1,3,5]

Answer 4

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 5

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.