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[英]How do I create a function parameter (variable) for the answers to an inquirer.prompt question in javascript? MySQL is also involved
[英]Inquirer.prompt exiting without an answer
我想知道是否有人可以幫助我解釋為什么下面的代碼會導致命令行退出而不等待用戶的回答。
init();
function init() {
loadPrompts();
}
async function loadPrompts() {
const { choice } = await inquirer.prompt([
{
type: "list",
name: "choice",
message: "What would you like to do?",
choices: [
{
name: "View All Employees",
value: "VIEW_EMPLOYEES",
},
{
name: "View All Employees By Department",
value: "VIEW_EMPLOYEES_BY_DEPARTMENT",
},
{
name: "View All Employees By Manager",
value: "VIEW_EMPLOYEES_BY_MANAGER",
},
{
name: "Add Employee",
value: "ADD_EMPLOYEE",
},
{
name: "Remove Employee",
value: "REMOVE_EMPLOYEE",
},
{
name: "Update Employee Role",
value: "UPDATE_EMPLOYEE_ROLE",
},
{
name: "Update Employee Manager",
value: "UPDATE_EMPLOYEE_MANAGER",
},
{
name: "View All Roles",
value: "VIEW_ROLES",
},
{
name: "Add Role",
value: "ADD_ROLE",
},
{
name: "Remove Role",
value: "REMOVE_ROLE",
},
{
name: "View All Departments",
value: "VIEW_DEPARTMENTS",
},
{
name: "Add Department",
value: "ADD_DEPARTMENT",
},
{
name: "Remove Department",
value: "REMOVE_DEPARTMENT",
},
{
name: "Quit",
value: "QUIT",
},
],
},
]);
switch (choice) {
case "VIEW_EMPLOYEES":
return viewEmployees();
default:
return quit();
}
}
async function viewEmployees() {
const employees = await db.findAllEmployees();
console.table(employees);
loadPrompts();
}
目的是一個簡單的命令行應用程序,它要求用戶 select 一個選項 - 然后根據他們選擇的 function 將被執行。 但是正在發生的是應用程序正在運行,顯示選項然后立即退出......
您應該將await
與loadPrompts()
一起使用以同步工作;
(async function init(){
await loadPrompts();
})();
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