c#重定向post請求

Question

嘗試使用第一個應用程序收到的相同表單數據參數從一個應用程序向另一個應用程序發出發布請求。 兩個應用程序控制器目前具有相同的方法：

public async Task<ActionResult> TestSet()
    {
        var inputString = Request.Form["inputString"];
        var inputFile = Request.Files[0];
        var resultString = await _service.Set(inputString, inputFile.FileName, inputFile.ContentType, inputFile.InputStream);
        return new MyJsonResult(new
        {
            fileName = resultString
        });
    }

返回json字符串：

{"fileName": "someFileName.png"}

試圖使第一種方法成為這樣的

public async Task<ActionResult> TestSet()
    {
        var inputString = Request.Form["inputString"];
        var inputFile = Request.Files[0];
        if (!string.IsNullOrEmpty(_redirectUrl))
        {
            using (var client = new HttpClient())
            {
                HttpContent content = GetContentSomehow(this.Request); // this i have an issue with
                var response = await client.PostAsync(_redirectUrl, content);
                var responseString = await response.Content.ReadAsStringAsync();
                return new MyJsonResult(responseString);
            }
        }
        var resultString = await _service.Set(inputString, inputFile.FileName, inputFile.ContentType, inputFile.InputStream);
        return new MyJsonResult(new
        {
            fileName = resultString
        });
    }

這可能有助於僅獲取 File 的 ByteArrayContent。

這可能可以將非文件參數放入 StringContent，但是如何將它們都放入單個 Content 中？

var jsonString = JsonConvert.SerializeObject(Request.Form.ToDictionary());
var content = new StringContent(jsonString, Encoding.UTF8, "application/json");

Answer 1

通過將 StringContent 和 StreamContent 合並到 MultipartFormDataContent 中解決了這個問題

public static MultipartFormDataContent GetMultipartFormData(HttpRequestBase req)
    {
        var formData = new MultipartFormDataContent();
        //formData.Headers.ContentType.MediaType = "multipart/form-data";
        foreach (var row in req.Form.ToDictionary())
        {
            formData.Add(new StringContent(row.Value), row.Key);
        }

        var file = req.Files[0];
        StreamContent fileStreamContent = new StreamContent(file.InputStream);
        fileStreamContent.Headers.ContentType = new MediaTypeHeaderValue(file.ContentType);
        formData.Add(fileStreamContent, file.FileName, file.FileName);

        return formData;
    }