Question

我有一个功能：

const Q = require('q');
const timers = require('timers');

function _testQ() {
    const deferred = Q.defer();
    console.log('test timer event 1s.');
    let t = timers.setInterval(() => {
        console.log('Timer event!');
        timers.clearInterval(t);
        deferred.resolve(true);
    }, 1000);
    return deferred.promise;
}
let out = _testQ().then((n) => {
    console.log('Debug1 n:', n);
});
console.log('Debug2 ', out);

如何以syncronios方式调用它。

我得到这样的输出：

test timer event 1s.
Debug2  { state: 'pending' }
Timer event!
Debug1 n: true

我需要它像这样打印：

test timer event 1 s.
Timer event!
Debug1 n: true
Debug2 {state: 'pending'}

Answer 1

如果你想在Debug1之后执行Debug2，你需要一个额外的.then块

let out = _testQ()
    .then((n) => {
        console.log('Debug1 n:', n);
    })
    .then(() => {
       console.log('Debug2 ');
    });

如何以同步方式调用Q函数

1 个答案: