asp.net核心Web API文件上传和“表单数据”多个参数传递给方法

时间:2018-08-17 09:47:56

标签: asp.net-mvc azure asp.net-core .net-core

我创建了一个以文件为参数的端点:

    [HttpPost("[action]")]
    [Consumes("multipart/form-data")]
    public ActionResult UploadImage(IFormFile  Files, string param)
    {

        long size = Files.Length;            
        var tempPath = Path.GetTempFileName();
        string file_Extension = Path.GetExtension(Files.FileName);                   
        var isValidFile = FileValidation.FileUploadValidation(Files);
        if (isValidFile.data)
        {
            string filename = Guid.NewGuid() + "" + file_Extension;
            return null;

        }
        else
        {
            return null;
        }
    }

我无法毫无问题地检索文件。 如何在同一方法中添加更多文本参数?

Debug View param parameter is null

Postmen call

4 个答案:

答案 0 :(得分:4)

它可以100%工作。经过测试。请执行以下步骤:

您应该为模型创建一个自定义类

public class FileInputModel
{
    public string Name { get; set; }
    public IFormFile FileToUpload { get; set; }
}

和类似的形式

<form method="post" enctype="multipart/form-data" asp-controller="Home" asp-action="UploadFileViaModel" >
    <input name="Name" class="form-control" />
    <input name="FileToUpload" type="file" class="form-control" />
    <input type="submit" value="Create" class="btn btn-default" />
</form>

和类似的控制器方法

[HttpPost]
public async Task<IActionResult> UploadFileViaModel([FromForm] FileInputModel model)
{
    if (model == null || model.FileToUpload == null || model.FileToUpload.Length == 0)
        return Content("file not selected");

    var path = Path.Combine(Directory.GetCurrentDirectory(), "wwwroot", model.FileToUpload.FileName);

    using (var stream = new FileStream(path, FileMode.Create))
    {
        await model.FileToUpload.CopyToAsync(stream);
    }

    return RedirectToAction("Files");
}

答案 1 :(得分:2)

[HttpPost("[action]")]
[Consumes("multipart/form-data")]
public IActionResult UploadImage([FromForm] FileInputModel Files)
{

    return Ok();
}

public class FileInputModel 
{
    public IFormFile File { get; set; }
    public string Param { get; set; }
}

在添加[FromForm]代码之后,需要在参数模型之前添加[FromForm]才能正常工作。

答案 2 :(得分:1)

我使用以下代码进行了测试,并且可以正常工作:

public class TestController : Controller
{
    [HttpPost("[controller]/[action]")]
    public IActionResult Upload(Model model)
    {
        return Ok();
    }

    public class Model
    {
        public IFormFile File { get; set; }
        public string Param { get; set; }
    }
}

请注意,您需要使用模型。

以下是具有相同属性的邮递员屏幕截图。

Postman screenshot below

更新:多个文件:

将模型更改为:

public class Model
{
    public List<IFormFile> Files { get; set; }
    public string Param { get; set; }
}

邮递员屏幕截图: Multipart form content

更新2

内容类型标题为multipart/form-data

以下是其工作的屏幕截图:

Working Code

答案 3 :(得分:1)

C#.NET Core

[HttpPost("someId/{someId}/someString/{someString}")]
[ProducesResponseType(typeof(List<SomeResponse>), StatusCodes.Status200OK)]
[ProducesResponseType(StatusCodes.Status401Unauthorized)]
[ProducesResponseType(StatusCodes.Status403Forbidden)]
[ProducesResponseType(typeof(Dictionary<string, string[]>), StatusCodes.Status400BadRequest)]
public async Task<IActionResult> Post(int someId, string someString, [FromForm] IList<IFormFile> files)
{
...
}

打字稿:

create(someId: number, someString: string, files: File[]) {
    var url = `${this.getBaseUrl()}/someId/${someId}/someString/${someString}`;

    const form = new FormData();
    for (let file of files) {
        form.append("files", file);
    }

    return this.http.post<SomeResponse>(url, form, {
        reportProgress: true,
    });
}