Require template literals instead of string concatenation JS-0246
Anti-pattern
Minor
a year ago
—
a year old
Unexpected string concatenation
10};
11
12export const app = {
13 view_ssh_keypair: baseURL + "/view/"14}Unexpected string concatenation
6 login: apiURL + "/api/v1/user/login",
7 create_ssh_keypair: apiURL + "/api/v1/key",
8 delete_ssh_keypair: apiURL + "/api/v1/key/",
9 logout: apiURL + "/api/v1/user/logout",10};
11
12export const app = {Unexpected string concatenation
5 change_password: apiURL + "/api/v1/user/change-password",
6 login: apiURL + "/api/v1/user/login",
7 create_ssh_keypair: apiURL + "/api/v1/key",
8 delete_ssh_keypair: apiURL + "/api/v1/key/", 9 logout: apiURL + "/api/v1/user/logout",
10};
11Unexpected string concatenation
4 retrieve_all_ssh_keypairs: apiURL + "/api/v1/key",
5 change_password: apiURL + "/api/v1/user/change-password",
6 login: apiURL + "/api/v1/user/login",
7 create_ssh_keypair: apiURL + "/api/v1/key", 8 delete_ssh_keypair: apiURL + "/api/v1/key/",
9 logout: apiURL + "/api/v1/user/logout",
10};Unexpected string concatenation
3export const api = {
4 retrieve_all_ssh_keypairs: apiURL + "/api/v1/key",
5 change_password: apiURL + "/api/v1/user/change-password",
6 login: apiURL + "/api/v1/user/login", 7 create_ssh_keypair: apiURL + "/api/v1/key",
8 delete_ssh_keypair: apiURL + "/api/v1/key/",
9 logout: apiURL + "/api/v1/user/logout",
Description
In ES2015 (ES6), we can use template literals instead of string concatenation.
var str = `Hello, ${name}!`;
Bad Practice
var str = "Hello, " + name + "!";
var str = "Time: " + (12 * 60 * 60 * 1000);
Recommended
var str = "Hello World!";
var str = `Hello, ${name}!`;
var str = `Time: ${12 * 60 * 60 * 1000}`;
var str = "Hello, " + "World!";