如何保持http通话一段时间? [英] How can I hold on http calls for a while?

问题描述

我遇到以下问题，如果有人可以给我发送一个主意，我已经尝试了一些，但没有用，我将不胜感激. 考虑代码:

I have the following problem and I would appreciate if someone could send me an idea, I have tried some, but it did not work. Consider the code:

 while (this.fastaSample.length > 0) {
      this.datainputService
        .saveToMongoDB(this.fastaSample.slice(0, batch))
        .subscribe();
    }

它想解决的问题是我无法在单个http调用中发送数据，因为它太大了，所以我能够发送10％的数据而没有任何问题，而且还不行！ 所以我想，我应该发送较小的批次，并且我已经在这里咨询过one Q& A，他们为我提供了帮助，但是并没有解决问题.

It supposes to solve the issue that I cannot send my data in a single http call, since it is too big, I was able to send 10% without issue, more than that, it does not work! So I thought, I should send smaller batches, and I have consulted sone Q&As here, and they helped me, but did not solve the problem.

我试图像在节点中那样使用await，但是它不起作用；它会一次发送所有的http，最好停止/保留代码，直到最后一个http调用完成，这将是一个不错的选择！ 有什么建议吗?

I have tried to use await as I did in node, but it does not work; it sends all the http at once, it would be nice to stop/hold the code until the last http call is complete, that would be nice! Any suggestion?

推荐答案

我想您可以通过使用from和concatAll将其与rxjs完美结合:

I suppose you could make it all nice and rxjs by using from and concatAll:

_{未测试的代码}

// first create batches by chunking the array
const batches = Array.from(
  { length: Math.ceil(fastaSample.length / batch) },
  (v, i) => fastaSample.slice(i * batch, i * batch + batch)
)

// Second go over these chunks using `from` and `concatAll`:
from(batches).pipe(
  map((batch) => this.data.inputService.saveToMongoDB(batch)),
  concatAll()
).subscribe();

这将连续拨打电话.如果可以同时执行请求，则可以执行mergeAll().

This will make the calls consecutively. If it's possible to do the requests at the same time, you can do mergeAll().

但是就像@Mike所评论的那样，似乎该问题应在MongoDB后端中处理并接受多部分请求.这样一来，您无需对内容进行分块

But like @Mike commented, it seems like the issue should be handled in the MongoDB backend and accept a multipart request. This way you don't need to chunk stuff

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