JavaScript如何替换函数从传播语法获得值?

Question

我有以下React代码，

const useDynamicReplaceVariable = ({ label, formikValues, fieldsTypes }) => {
const { locale } = useTranslationState();
const formattedLabel = useMemo(() => {
const variablesRegex = /?([w-]+)?/g;
let labelResult = label;

if (variablesRegex.test(label)) {
labelResult = ' ';
if (Object.keys(formikValues).length > 0) {
labelResult = label.replace(variablesRegex, (_, ...[variable]) => {
const type = fieldsTypes[variable];
// Set undefined or null to empty
let variableValue = (formikValues[variable] === undefined || formikValues[variable] == null) ? '' : formikValues[variable];
if (variableValue && [DesignerDdTypes.DatePicker, DesignerDdTypes.DatetimePicker].includes(type)) {
variableValue = dateToString(variableValue, locale, type === DesignerDdTypes.DatetimePicker);
}
return variableValue;
});
}
}
return labelResult;
// eslint-disable-next-line react-hooks/exhaustive-deps  
}, [label, JSON.stringify(formikValues)]);
return { formattedLabel };
};

我看不懂labelResult = label.replace(variablesRegex, (_, ...[variable])这行，没有定义variable，为什么要用扩展语法?

Answer 1

...[variable]是

的简写

function someFunction(...args) {
const [variable] = args
}

someFunction(1, 2, 3, 4, 5)
function someFunction(arg1, ...args) {
console.log('arg1', arg1)
console.log('args', args)
const [variable] = args
console.log('variable', variable)
}

someFunction(1, 2, 3, 4, 5)
function someFunction(arg1, ...[variable]) {
console.log('arg1', arg1)
console.log('variable', variable)
}