Function to emulate the ES7 destructuring with spread operator behaviour in Typescript

destructure.ts

/**
 * Return a copy of `obj` with the specified properties removed
 * Modified from Babel source code
 */
function objectWithoutProperties (obj: Object, keys: string[]): any {
    let target = {}
    for (var i in obj) {
        if (keys.indexOf(i) >= 0) { continue }
        if (!Object.prototype.hasOwnProperty.call(obj, i)) { continue }
        target[i] = obj[i];
    }
    return target;
}

/**
 * Function to emulate the ES7 destructuring with spread operator behaviour, commonly used with Redux.
 * Returns a copy of `obj` with the properties specified in `keys` copied over,
 * and all remaining properties inside an object in a property called `__rest`.
 *
 * Example:
 * ES7 code:      const { value, onChange, onBlur, ...props } = this.props
 * ES6 with this: const { value, onChange, onBlur, __rest: props } = destructure(this.props, ['value', 'onChange', 'onBlur'])
 *
 * TODO deprecate this when tsc supports the ES7 operator: https://github.com/Microsoft/TypeScript/issues/2103
 */
export function destructure(obj: Object, keys: string[]): any {
  let destructured: any = {}
  // Map all the properties in keys on to the new object
  for (let key of keys) {
    destructured[key] = obj[key]
  }
  // And then get the left over properties and assign to "__rest"
  destructured.__rest = objectWithoutProperties(obj, keys)
  return destructured
}

Can you give a usage example?

@amcdnl

const { value, onChange, onBlur, __rest: props } = destructure(this.props, ['value', 'onChange', 'onBlur'])

not so perfect :) I'll wait