是否有干净/短/正确的方式共同使用axios承诺并上传进度事件?
假设我有下一个上传功能:
function upload(payload, onProgress) {
const url = '/sources/upload';
const data = new FormData();
data.append('source', payload.file, payload.file.name);
const config = {
onUploadProgress: onProgress,
withCredentials: true
};
return axios.post(url, data, config);
}
此函数返回了承诺。
我也有一个传奇:
function* uploadSaga(action) {
try {
const response = yield call(upload, payload, [?? anyProgressFunction ??]);
yield put({ type: UPLOADING_SUCCESS, payload: response });
} catch (err) {
yield put({ type: UPLOADING_FAIL, payload: err });
}
}
我希望接收进度事件并将其传递给saga。此外,我想捕获axios请求的成功(或失败)结果。有可能吗?
感谢。
答案 0 :(得分:7)
所以我找到了答案,感谢Mateusz Burzyński澄清。
我们需要使用eventChannel,但有点刺激。
假设我们有上传文件的api功能:
function upload(payload, onProgress) {
const url = '/sources/upload';
const data = new FormData();
data.append('source', payload.file, payload.file.name);
const config = {
onUploadProgress: onProgress,
withCredentials: true
};
return axios.post(url, data, config);
}
在saga中,我们需要创建eventChannel,但将emit放在外面。
function createUploader(payload) {
let emit;
const chan = eventEmitter(emitter => {
emit = emitter;
return () => {}; // it's necessarily. event channel should
// return unsubscribe function. In our case
// it's empty function
});
const uploadPromise = upload(payload, (event) => {
if (event.loaded.total === 1) {
emit(END);
}
emit(event.loaded.total);
});
return [ uploadPromise, chan ];
}
function* watchOnProgress(chan) {
while (true) {
const data = yield take(chan);
yield put({ type: 'PROGRESS', payload: data });
}
}
function* uploadSource(action) {
const [ uploadPromise, chan ] = createUploader(action.payload);
yield fork(watchOnProgress, chan);
try {
const result = yield call(() => uploadPromise);
put({ type: 'SUCCESS', payload: result });
} catch (err) {
put({ type: 'ERROR', payload: err });
}
}