How to Preserve the Root Object Name When Destructuring Function Parameters in ES6?

ES6 Destructuring Function Parameter: Preserving Root Object Name

In ES6, destructuring function arguments allows for concise and expressive parameter handling. However, when working with multiple configuration parameters spread across a parent-child class hierarchy, the name of the root object may be lost after destructuring.

In ES5, you can pass the root object up the hierarchy using inheritance:

// ES5
var setupParentClass5 = function(options) {
    textEditor.setup(options.rows, options.columns);
};

var setupChildClass5 = function(options) {
    rangeSlider.setup(options.minVal, options.maxVal);
    setupParentClass5(options); // Pass the options object up
};

However, in ES6 using destructuring, this becomes an issue:

// ES6
var setupParentClass6 = ({rows, columns}) => {
    textEditor.setup(rows, columns);
};

var setupChildClass6 = ({minVal, maxVal}) => {
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6( /* ??? */ );  // How to pass the root options object?
};

To retain the root object's name, you can declare it as a parameter before destructuring:

const setupChildClass6 = options => {
    const {minVal, maxVal} = options;
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6(options);
};

Alternatively, you can extract all options individually in setupChildClass6:

var setupChildClass6b = ({minVal, maxVal, rows, columns}) => {
    rangeSlider.setup(minVal, maxVal);
    setupParentClass6({rows, columns});
};

However, this approach requires more lines of code and may not be optimal when dealing with multiple destructured variables.

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