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[英]Promise to run nested promises sequentially and resolve upon first reject
[英]Promises can't run code after resolve/reject
我在異步功能中有一個異步功能。 在第二個步驟中,我必須等待promise解決或拒絕時以及在運行下面的其他代碼之后。 但是如果諾言拒絕我的代碼停止並且不運行其他功能。 我該如何解決?
await axios.all(promises).then(res => {
axios.patch("/url", { foo: bar }).then(async () => {
const promises2 = arr.map(item => {
return axios.post("/url-2", item)
});
await Promise.all(promises2)
.then(() => console.log("resolved")) //this not calling ever
.catch(() => console.log("failed")) //this not calling ever
console.log("This console log ever not working")
})
})
承諾未正確鏈接, axios.patch(...)
承諾未返回。 await
then
catch
語法糖,其目的是在可能的情況下擺脫嵌套函數。 它應該是:
const res = await axios.all(promises)
await axios.patch("/url", { foo: bar })
const promises2 = arr.map(item => {
return axios.post("/url-2", item)
});
try {
await Promise.all(promises2)
console.log("resolved"))
} catch (err) {
console.log("failed");
}
您的代碼順序錯誤。 當然,如果第一個承諾被拒絕,那么其余的將不會被調用。
嘗試通過以下方式重寫代碼:
let res = await axios.all(promises).catch(() => { console.log("failed"); return false; });
if (!res) {
// Do something when rejected
....
}
// Call the 2nd promise
let res2 = await axios.path("/url", {foo: bar}).catch(() => {console.log("failed 2"); return false; });
if (!res2) {
// Do something when the 2nd promise is rejected
...
}
// Call your last promise
let res3 = await Promise.all(promises2).catch(() => {console.log("failed 3"); return false; });
if (!res3) {
// Do something if it is rejected again
....
}
// Otherwise, do your thing
試試這個代碼,它應該指出錯誤或拒絕發生的位置(即肯定是在Promise.all(promises2)
運行之前
await axios.all(promises)
.then(res => axios.patch("/url", { foo: bar }), err => {
throw `all(promises) failed with ${err}`;
})
.then(() => {
const promises2 = arr.map(item => {
return axios.post("/url-2", item);
});
return Promise.all(promises2)
.then(() => console.log("resolved")) //this not calling ever
.catch(err => {
throw `all(promises2) failed with ${err}`;
});
}, err => {
throw `patch failed with ${err}`;
})
.catch(err => console.error(err));
注意,我已經刪除了async / await,因為在您發布的代碼中完全沒有必要
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