如何在 ASP.NET Core 中模拟 IFormFile 以进行单元/集成测试？

Question

我想编写用于在 ASP.NET Core 中上传文件的测试，但似乎找不到模拟/实例化从 IFormFile 派生的对象的好方法。

关于如何做到这一点的任何建议？

Answer 1

假设你有一个像..这样的控制器

public class MyController : Controller {
    public Task<IActionResult> UploadSingle(IFormFile file) {...}
}

...IFormFile.OpenReadStream() 是通过被测方法访问的。

您可以使用Moq 模拟框架创建一个测试来模拟流数据。

[TestClass]
public class IFormFileUnitTests {
    [TestMethod]
    public async Task Should_Upload_Single_File() {
        //Arrange
        var fileMock = new Mock<IFormFile>();
        //Setup mock file using a memory stream
        var content = "Hello World from a Fake File";
        var fileName = "test.pdf";
        var ms = new MemoryStream();
        var writer = new StreamWriter(ms);
        writer.Write(content);
        writer.Flush();
        ms.Position = 0;
        fileMock.Setup(_ => _.OpenReadStream()).Returns(ms);
        fileMock.Setup(_ => _.FileName).Returns(fileName);
        fileMock.Setup(_ => _.Length).Returns(ms.Length);

        var sut = new MyController();
        var file = fileMock.Object;

        //Act
        var result = await sut.UploadSingle(file);

        //Assert
        Assert.IsInstanceOfType(result, typeof(IActionResult));
    }
}

或者，从 ASP.NET Core 3.0 开始，使用 FormFile Class 的实例，它现在是 IFormFile 的默认实现。

这是使用FormFile 类进行上述相同测试的示例

[TestClass]
public class IFormFileUnitTests {
    [TestMethod]
    public async Task Should_Upload_Single_File() {
        //Arrange
       
        //Setup mock file using a memory stream
        var content = "Hello World from a Fake File";
        var fileName = "test.pdf";
        var stream = new MemoryStream();
        var writer = new StreamWriter(stream);
        writer.Write(content);
        writer.Flush();
        stream.Position = 0;

        //create FormFile with desired data
        IFormFile file = new FormFile(stream, 0, stream.Length, "id_from_form", fileName);
        
        MyController sut = new MyController();
        
        //Act
        var result = await sut.UploadSingle(file);

        //Assert
        Assert.IsInstanceOfType(result, typeof(IActionResult));
    }
}

Answer 2

创建一个实际的内存实例会更容易

 var bytes = Encoding.UTF8.GetBytes("This is a dummy file");
 IFormFile file = new FormFile(new MemoryStream(bytes), 0, bytes.Length, "Data", "dummy.txt");

Answer 3

添加 harishr 的答案并设置长度（这是我的 Blob.Upload() 工作所需要的）。

private IFormFile CreateTestFormFile(string fileName, string content)
{
    byte[] bytes = Encoding.UTF8.GetBytes(content);

    return new FormFile(
        baseStream: new MemoryStream(bytes),
        baseStreamOffset: 0,
        length: bytes.Length,
        name: "Data",
        fileName: fileName
    );
}

Answer 4

我无法模拟 IFormFile，因为我遇到了 Stream 问题，所以我最终创建了 FormFile 的实例。如果您对文件使用验证器等，请注意您需要设置内容类型。

        private IFormFile GetFileMock(string contentType, string content)
        {
            byte[] bytes = Encoding.UTF8.GetBytes(content);

            var file = new FormFile(
                baseStream: new MemoryStream(bytes),
                baseStreamOffset: 0,
                length: bytes.Length,
                name: "Data",
                fileName: "dummy.csv"
            )
            {
                Headers = new HeaderDictionary(),
                ContentType = contentType
            };

            return file;
        }
        

打电话

 GetFileMock("text/csv", "test;test;")

Answer 5

private static IFormFileCollection GetFormFileCollection()
{
    var filesFolder = $"{AppDomain.CurrentDomain.SetupInformation.ApplicationBase}UploadFiles\\";
    List<string> filesPathsListToUpload = new List<string>();
    filesPathsListToUpload.Add($"{filesFolder}UploadFile1.png");
    filesPathsListToUpload.Add($"{filesFolder}UploadFile2.jpg");
    filesPathsListToUpload.Add($"{filesFolder}UploadFile3.bmp");

    FormFileCollection filesCollection = new FormFileCollection();
    foreach (var filePath in filesPathsListToUpload)
    {
        var stream = File.OpenRead(filePath);
        IFormFile file = new FormFile(stream, 0, stream.Length, "files", Path.GetFileName(filePath))
        {
            Headers = new HeaderDictionary(),
            ContentType = filePath.Split('.')[1] == "jpg" ? "image/jpeg"
                : filePath.Split('.')[1] == "png" ? "image/png"
                : "image/bmp",
        };
        
        filesCollection.Add(file);
    }

    var httpContext = new DefaultHttpContext();
    httpContext.Request.Headers.Add("Content-Type", "multipart/form-data");
    httpContext.Request.Form = new FormCollection(new Dictionary<string, StringValues>(), filesCollection);

    return httpContext.Request.Form.Files;
}