Javascript - Apply trim function to each string in an array

Question

Want to trim each string in an array, e.g., given

x = [\' aa \', \' bb \'];

output

[\'aa\', \'bb\']

My fi

Answer 1

Or this can be solved with arrow functions:

x.map(s => s.trim());

Answer 2

Another ES6 alternative

const row_arr = ['a ', ' b' , ' c ', 'd'];
const trimed_arr = row_arr.map(str => str.trim());
console.log(trimed_arr); // <== ['a', 'b', 'c', 'd']

Answer 3

If you are using JQuery, then a better way to do this, as it will work with IE8 as well (I need to support IE8) is this:

$.map([' aa ', ' bb ', '   cc '], $.trim);

Answer 4

The simple variant without dependencies:

 for (var i = 0; i < array.length; i++) {
     array[i] = array[i].trim()
 }

ES6 variant:

const newArray = oldArray.map(string => string.trim())

ES6 function variant:

const trimmedArray = array => array.map(string => string.trim())

Answer 5

String.prototype.trim.apply is the Function.prototype.apply method without being bound to trim. map will invoke it with the string, the index and the array as arguments and nothing (undefined) for the thisArg - however, apply expects to be called on functions:

var apply = String.prototype.trim.apply;
apply.call(undefined, x[0], 0, x) // TypeError

What you can do is passing the trim function as the context for call:

[' aa ', ' bb '].map(Function.prototype.call, String.prototype.trim)
// ['aa', 'bb']

What happens here is

var call = Function.prototype.call,
    trim = String.prototype.trim;
call.call(trim, x[0], 0, x) ≡
      trim.call(x[0], 0, x) ≡
            x[0].trim(0, x); // the arguments don't matter to trim