我有一个n个可观察的数组，我想一次发出4个请求，然后再发出4个，直到最后

Question

如何限制并行请求？

for(let offset=0；offset<file[0].size；offset+=chunkSize({const chunk=file[0].slice(offset，offset+chunkSize(；

const blockId = String(Math.floor(100000 + Math.random() * 900000));
const base64BlockId = btoa(blockId);
this.blockids.push(base64BlockId);
request.push(this.blobService.addNewVideo(uniqueName, chunk, base64BlockId));
}
forkJoin(
request.map(req => req.pipe(
tap((event: HttpEvent<any>) => {
switch (event.type) {
case HttpEventType.UploadProgress:
this.progress = Math.round((event.loaded / event.total!) * 100);
// console.log(`uploaded! ${this.progress}%`);
break;
case HttpEventType.Response:
console.log('Successfully uploaded', event.body," : ",event.status,event.statusText);
}
})
))
).subscribe(() => {
this.commit(uniqueName);
});

Answer 1

您可以将mergeAll用于并发请求。

参考编号：https://levelup.gitconnected.com/limit-concurrency-with-rxjs-and-promises-78590d2c85d0