async/await怎么抛出异常
var method = return new Promise((resolve, reject) =>{。。。。。} var f1 = async function() {await method}， var f2 = async function() { f1() } 如果不在f1的await method这里进行try/catch，那么怎么把异常继续往上抛，让f2来进行try/catch 直接在f2进行try/catch会报UnhandledPromiseRejectionWarning
4 回复
async function f2() {
try {
await f1();
} catch {
console.log('catch!');
}
}
只有用await时才会被try/catch捕获，否则它就是一个普通的promise
async function f2() {
f1().catch(() => console.log('catch!'));
}
