如何在Xunit的Catch块中编写单元测试？

Question

public IDictionary Build（string content）{尝试{var settings = new JsonSerializerSettings{// NullValueHandling = NullValueHandling。忽略，MissingMemberHandling = MissingMemberHandling.Ignore};var contentStudioResponse = JsonConvert.DeserializeObject>（内容，设置）；

            if (contentStudioResponse?.Items == null)
            {
                _logger.Warning("No records found in content studio response for label:({@content})", content);
                return new Dictionary<string, Label>();

            }

            return contentStudioResponse.Items.ToDictionary(x => x.Key,
                x => new Label
                {
                    Value = x.DynamicProperties.MicroContentValue
                }
            );
        }
        catch (Exception e)
        {
            _logger.Error(e, "Failed to deserialize or build contentstudio response for label");
            return new Dictionary<string, Label>();
        }


//Below is my solution which is not working.
    [Fact]
    public void Builder_ThrowsException()
    {
        string json_responsive_labels = "abcd";
        var builder = new LabelBuilder(_testLogger).Build(json_responsive_labels);
        Assert.Throws<Exception>(() => builder);
        //var sut = new LabelBuilder(_testLogger);            
        //Should.Throw<Exception>(() => sut.Build(json_responsive_labels));
    }

Answer 1

具有this的通读。这一步一步地说明了如何测试是否引发了异常。

但是，根据您所写的内容，代码不会引发异常，因为此时您只记录了异常，然后返回Dictionary。

   catch (Exception e)
   {
      _logger.Error(e, "Failed to deserialize or build contentstudio response for label");
      return new Dictionary<string, Label>();
   }

您实际上想要做的是明确地抛出一个异常，如下所示：

   catch (Exception e)
   {
      throw new Exception();
   }

这样做，您的代码将引发一个异常，您可以捕获该异常并对其进行测试。