异步等待模式帮助.我做得对吗？

Question

我有一个"api"例如存储库模式,我写了从web返回xml然后将它水合成类.

它似乎挂起了GetXmlAsync(url)方法的结果.

public async Task<string> GetXmlAsync(string url)
{
    string xml = string.Empty;

    HttpMessageHandler handler = new HttpClientHandler();
    HttpClient httpClient = new HttpClient(handler);
    Uri uri = new Uri(url, UriKind.Absolute);
    HttpResponseMessage response = await httpClient.GetAsync(uri);

    xml = await response.Content.ReadAsStringAsync();

    return xml;
}

但是当我在单元测试中使用相同的代码时,它可以工作.

在应用程序中,我称之为:

public async Task<IEnumerable<Post>> GetRecentAsync(int page)
{
    string url = this.urls.GetRecent(page);
    string xml = string.Empty;

    var xmlTask = this.loader.GetXmlAsync(url);

    xml = xmlTask.Result; // Hangs right here.

    var results = this.modelLoader.XmlToPost(xml);

    if (results.Count() < 1)
    {
        this.LastError = XmlLoadError;
    }

    return results.AsEnumerable();
}

[TestMethod]
public async Task Integration_HttpLoader_GetXmlAsync_GetRecent_Xml_ShouldNotBeNullOrEmpty()
{
    int page = 1;
    string xml = string.Empty;
    IUrl url = this.GetUrlHelper();
    ILoader loader = this.GetIntegrationLoader(false);

    xml = await loader.GetXmlAsync(url.GetRecent(page));

    Assert.IsTrue(!string.IsNullOrEmpty(xml));
}

Answer 1

在您的应用程序中,您没有在等待调用this.loader.GetXmlAsync(url)之前等待您点击此行,在另一个线程上触发异步任务,然后立即前进到下一行而没有得到响应.它适用于您的单元测试,因为您正确使用了await关键字.