Question

我有一个简单的集成测试。

        [TestMethod]
    public async Task INT_GetSomething_Sucess()
    {
        //Arrange
        HttpResponseMessage response;
        BidMonthDetails returnedObj;
        //Act
        try
        {
            string request = JsonConvert.SerializeObject(DateTime.Now);
            response = await TestClient.PostAsync("/api/Trade/GetSomething", new StringContent(request, Encoding.UTF8, "application/json"));
            var jsonString = await response.Content.ReadAsStringAsync();
            returnedObj = JsonConvert.DeserializeObject<MyModel>(jsonString);
        }
        catch (Exception ex)
        {

            throw ex;
        }


        //Assert
        Assert.AreEqual(response.StatusCode, System.Net.HttpStatusCode.OK);
        Assert.IsNotNull(returnedObj);


    }

此测试在贸易控制器中称为GetSomething的功能

     [HttpPost]
    public async Task<ActionResult<MyModel>> GetSomething(DateTime date)
    {

在这一点上，“日期”仅是最小日期。我希望它是今天的日期。

Answer 1

默认情况下，假设DateTime参数来自查询字符串。由于您没有date查询字符串参数，因此它默认为DateTime的默认值DateTime.Min

要解决此问题，您有两种选择：

1）用FromBody属性装饰参数：

public async Task<ActionResult<MyModel>> GetSomething([FromBody]DateTime date)

2）将值放在查询字符串而不是正文中：

response = await TestClient.PostAsync($"/api/Trade/GetSomething?date={DateTime.UtcNow}"...

Asp.net Core的POST方法在模型绑定上将DateTime.Now转换为Min Date

1 个答案: