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js_task_2 task_1 task_2 #20

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MikhailBobrov
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const typeOfPhone_2 = new RegExp("^\\+7\\d{10}$");
const typeOfEmail = new RegExp("^\\w+@\\w+\\.\\w+$");

let item = {}
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item нигде не переприсвается, лучше в таких случаях использовать const вместо let

@@ -20,6 +20,50 @@
@returns {boolean} Результат добавления
*/
function add(phoneBook, phone, name, email) {
const typeOfPhone_1 = new RegExp("^\\+7\\-\\d{3}\\-\\d{3}\\-\\d{2}\\-\\d{2}$");
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Следует придерживаться одного стиля именования переменных, для js это предпочтительно camelCase, а у тебя тут смешанный camelCase и snake_case.

@@ -20,6 +20,50 @@
@returns {boolean} Результат добавления
*/
function add(phoneBook, phone, name, email) {
const typeOfPhone_1 = new RegExp("^\\+7\\-\\d{3}\\-\\d{3}\\-\\d{2}\\-\\d{2}$");
const typeOfPhone_2 = new RegExp("^\\+7\\d{10}$");
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Аналогично комменту выше


let item = {}

if (phone.match(typeOfPhone_1)) {
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метод match возвращает сопоставленное значение, ты его нигде не используешь, тут будет уместнее использовать метод test() у регулярки, который возвращает bool

if (phone.match(typeOfPhone_1)) {
item.phone = phone;
}
else if(phone.match(typeOfPhone_2)) {
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Можно объеденить с предыдущим ifом, если написать через логический оператор ИЛИ

function find(phoneBook, query) {
if (query === '*') {
let firstAr = [];
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const. Почему бы не сделать один общий массив результатов, и возвращать его, нежели каждый раз плодить новый

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ок, сделал


let oldFormatPhone = phoneBook[key].phone; // +7-922-555-35-35

let newOne = oldFormatPhone.replace(/(\+\d)(\-?)(\d{3})(\-?)(\d{3})(\-?)(\d{2})(\-?)(\d{2})/,
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Вынести повторяющуюся функциональности в 3х ifах в отдельную функицю

}

}
return thirdAr
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А где сортировка по имени?

return array_1.length;
}

else if (query === 'andrey') {
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Используй написанный метод find в предыдущем задании, а здесь напиши только удаление

if (item.match(typeOfPhone_1)) {
objItem.phone = item;
}
else if (item.match(typeOfPhone_2)) {
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Комменты из 1го задания актуальны

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3 participants