c# - 链接 TransformBlock 产生 IEnumerable阻止接收 T

Question

我正在编写一个网络画廊刮板，我想尽可能地使用 TPL Dataflow 并行处理文件。

为了抓取，我首先获取画廊主页并解析 HTML 以获取图像页面链接作为列表。然后我转到列表中的每个页面并解析 HTML 以获取指向图像的链接，然后将其保存到磁盘。

这是我的程序的大纲：

var galleryBlock = new TransformBlock<Uri, IEnumerable<Uri>>(async uri =>
{
    // 1. Get the page
    // 2. Parse the page to get the urls of each image page
    return imagePageLinks;
});

var imageBlock = new TransformBlock<Uri, Uri>(async uri =>
{
    // 1. Go to the url and fetch the image page html
    // 2. Parse the html to retrieve the image url
    return imageUri;
});

var downloadBlock = ActionBlock<Uri>(async uri =>
{
    // Download the image from uri to list
});

var opts = new DataflowLinkOptions { PropagateCompletion = true};
galleryBlock.LinkTo(imageBlock, opts); // this doesn't work, as I'm returning a list and not a single Item. However I want to progress that block in parallel.
imageBlock.LinkTo(downloadBlock, opts);

标签： c#asynchronousparallel-processingtpl-dataflow

Answer 1

您可以使用 aTransformManyBlock代替您的TransformBlock：

var galleryBlock = new TransformManyBlock<Uri, Uri>(async uri =>
{
    return Enumerable.Empty<Uri>(); //just to get it compiling
});

var imageBlock = new TransformBlock<Uri, Uri>(async uri =>
{
    return null;  //just to get it compiling
});

var opts = new DataflowLinkOptions { PropagateCompletion = true };
galleryBlock.LinkTo(imageBlock, opts); // bingo!