function test() {
return new Promise((resolve, reject)=>{
resolve(123);
});
}
test().then((a)=>{
console.log(a);
}).then(function (b) {
console.log(b);//这行代码被执行了
});The first parameter method of then should be that promise is executed when resolved, but console.log(b) is actually executed. Why is this
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An uninvited reply!
In Promise,
.then( func )会返回一个新的 Promise 实例，这个实例的结果，是把函数体 func 的结果通过执行Promise.resolve()得到的。所以在你的问题里，console.log(a)没有返回值，相当于Promise.resolve(null)得到一个状态为resolvedis a Promise instance, so it will continue to the next step.Regarding Promise, I recommend you to read my tutorial: N ways of using Promise. It has a very detailed explanation and can basically answer all questions about Promise.
.then(parameter 1, parameter 2) Parameter 1 is the success callback function (resolve()), parameter 2 is the failure callback function (reject()), because of the way you write it here, the .then after the second one (the first function), will be executed, and the value of b is printed out as undefined, that’s about it! !