c# - 使用 puppeteersharp 返回 Image base 64 而不将文件保存在硬盘中

Question

我正在使用 puppeteersharp 将 HTML 转换为图像：

https://www.puppeteersharp.com/examples/index.html

这是获取 HTMl 并保存图像，然后我将返回 base64 图像，我希望返回图像 base64 而不将其保存在硬盘中。

这是代码：

          await new BrowserFetcher().DownloadAsync(BrowserFetcher.DefaultRevision);
           var browser = await Puppeteer.LaunchAsync(new LaunchOptions
        {
            Headless = true
        });
        var page = await browser.NewPageAsync();

        await page.SetViewportAsync(new ViewPortOptions
        {
            Width = 750,
            Height = 750,
        });

        await page.SetContentAsync("<h2>Test</h2>");
        await page.ScreenshotAsync(@"/Users/myname/Documents/puppeteertests/test1.jpg");

        byte[] b = System.IO.File.ReadAllBytes(@"/Users/myname/Documents/puppeteertests/test1.jpg");
        return Convert.ToBase64String(b);

Answer 1

我在想这样的事情应该可以工作 - 它获取页面的“ScreenshotStreamAsync”方法返回的 Stream 并将其复制到 MemoryStream 以便我们可以非常简单地从中获取一个字节数组并传递该字节数组到 Convert.ToBase64String

var thing =  await new BrowserFetcher().DownloadAsync(BrowserFetcher.DefaultRevision);
        var browser = await Puppeteer.LaunchAsync(new LaunchOptions
        {
            Headless = false
        });
        var page = await browser.NewPageAsync();

        await page.SetViewportAsync(new ViewPortOptions
        {
            Width = 750,
            Height = 750,
        });

        await page.SetContentAsync("<h2>Test</h2>");
        await page.ScreenshotAsync(@"test1.jpg");

        using (var memStream = new MemoryStream())
        {
           page.ScreenshotStreamAsync().Result.CopyTo(memStream);
           var mything = Convert.ToBase64String(memStream.ToArray());
           Console.WriteLine(mything);
        }