.Net C#向远程服务器Api上传文件

时间:2020-04-13
本文章向大家介绍.Net C#向远程服务器Api上传文件,主要包括.Net C#向远程服务器Api上传文件使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

Api服务代码一:

        /// <summary>
        /// 服务器接收接口
        /// </summary>
        [HttpPost]
        [Route("ReceiveFile")]
        public HttpResponseMessage ReceiveFile()
        {
            string result = string.Empty;
            ArrayList list = new ArrayList();
            try
            {
                Stream postStream = HttpContext.Current.Request.InputStream;
                byte[] b = new byte[postStream.Length];

                string postFileName = DNTRequest.GetString("fileName");
                if (string.IsNullOrEmpty(postFileName) && HttpContext.Current.Request["fileName"] != null)
                {
                    postFileName = HttpContext.Current.Request["fileName"];
                }
                string fileExtension = Path.GetExtension(postFileName);
                string dirName = "other";
                if (!string.IsNullOrEmpty(fileExtension))
                {
                    dirName = fileExtension.Substring(fileExtension.LastIndexOf(".") + 1);
                }
                string dir = "/_temp/file/" + dirName + "/" + DateTime.Now.ToString("yyyyMMdd") + "/";
                string fileName = DateTime.Now.ToString("yyyyMMddHHmmss_ffff");
                fileName += fileExtension;
                string filePath = HttpContext.Current.Server.MapPath(dir);
                string saveFilePath = Path.Combine(filePath, fileName);
                string dirPath = dir + fileName;
                list.Add(dirPath);

                if (!Directory.Exists(filePath))
                {
                    Directory.CreateDirectory(filePath);
                }

                FileStream fs = new FileStream(saveFilePath, FileMode.Create);
                byte[] new_b = new byte[1024];
                const int rbuffer = 1024;

                while (postStream.Read(new_b, 0, rbuffer) != 0)
                {
                    fs.Write(new_b, 0, rbuffer);
                }
                postStream.Close();
                fs.Close();
                fs.Dispose();

                if (list.Count > 0)
                {
                    result = DNTRequest.GetResultJson(true, "success", string.Join(",", list.ToArray()));
                }
            }
            catch (Exception ex)
            {
                result = DNTRequest.GetResultJson(false, ex.Message, null);
            }
            HttpResponseMessage responseMessage = new HttpResponseMessage { Content = new StringContent(result, Encoding.GetEncoding("UTF-8"), "text/plain") };
            return responseMessage;
        }
 

Api服务代码二:

 
        [HttpPost]
        [Route("ReceiveFileTest")]
        public HttpResponseMessage ReceiveFileTest()
        {
            string result = string.Empty;
            var request = HttpContext.Current.Request;
            try
            {
                if (request.Files.Count > 0)
                {
                    var fileNameList = new List<string>();
                    string dirName = "other";
                    foreach (string f in request.Files)
                    {
                        var file = request.Files[f];

                        string fileExtension = Path.GetExtension(file.FileName).ToLower();
                        string fileName = DateTime.Now.ToString("yyyyMMddHHmmss_ffff");
                        fileName += fileExtension;
                        if (!string.IsNullOrEmpty(fileExtension))
                        {
                            dirName = fileExtension.Substring(fileExtension.LastIndexOf(".") + 1);
                        }
                        string dir = "/_temp/file/" + dirName + "/" + DateTime.Now.ToString("yyyyMMdd") + "/";
                        string filePath = HttpContext.Current.Server.MapPath(dir);
                        if (!Directory.Exists(filePath))
                        {
                            Directory.CreateDirectory(filePath);
                        }
                        string fileSavePath = Path.Combine(filePath, fileName);

                        Stream postStream = file.InputStream;
                        //
                        FileStream fs = new FileStream(fileSavePath, FileMode.Create);
                        byte[] new_b = new byte[1024];
                        const int rbuffer = 1024;

                        while (postStream.Read(new_b, 0, rbuffer) != 0)
                        {
                            fs.Write(new_b, 0, rbuffer);
                        }
                        postStream.Close();
                        fs.Close();
                        fs.Dispose();

                        string dirPath = dir + fileName;
                        fileNameList.Add(dirPath);
                        fileNameList.Add(fileName);
                    }

                    result = DNTRequest.GetResultJson(true, string.Format("{0}:{1}", HttpStatusCode.OK, string.Join(",", fileNameList.ToArray())), null);
                }
                else
                {
                    result = DNTRequest.GetResultJson(false, "请选择上传的文件", null);
                }
            }
            catch (Exception ex)
            {
                result = DNTRequest.GetResultJson(false, ex.Message, null);
            }
            HttpResponseMessage responseMessage = new HttpResponseMessage { Content = new StringContent(result, Encoding.GetEncoding("UTF-8"), "text/plain") };
            return responseMessage;
        }
 

Ajax提交file代码,调用Api代码一的Script:

此方法后端Api代码一中始终获取不到文件名,所以我在Ajax的url中加了?fileName=fileObj.name,这样调试能走通。为什么Ajax提交时不能获取到文件名,我还在研究。

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<script>
 
        $("#inpSubmit").click(function () {
            //var fileObj = new FormData($("#importModel")[0]);
            var fileObj = document.getElementById("inpFileControl").files[0];
            if (typeof (fileObj) == "undefined" || fileObj.size <= 0) {
                alert("请选择文件");
                return;
            }
            //console.log(fileObj);
 
            var formFile = new FormData();
            formFile.append("fileName", fileObj.name);
            formFile.append("file", fileObj);
            //console.log(JSON.stringify(formFile));
 
            //var paramters = {};
            //paramters.timestamp = new Date().getTime();
            //paramters.fileName = fileObj.name;
            //paramters.file = fileObj;
            //console.log(JSON.stringify(paramters));
 
            $.ajax({
                type: "post",
                url: "http://localhost:19420/Api/ReceiveFile?fileName=" + fileObj.name,
                dataType: "json",
                //contentType: false,
                processData: false,//用于对data参数进行序列化处理,默认值是true。默认情况下发送的数据将被转换为对象,如果不希望把File转换,需要设置为false
                cache: false,
                data: fileObj,
                success: function (json) {
                    if (json.result) {
                        $("#inpFileUrl").val(json.data);
                        console.log(json.data);
                    }
                    else {
                        alert(json.msg);
                    }
                },
                error: function (ret) {
                    console.log(ret.responseText);
                }
            });
            return false;
        });
    </script>

Ajax提交file代码,调用Api代码一的Html:

 
<form>
    <fieldset>
        <legend>Ajax提交到远程服务器Api</legend>
        <div class="form-group">
            <label class="col-sm-2 control-label" for="ds_host">选择文件</label>
            <div class="col-sm-4">
                <input class="form-control" id="inpFileUrl" name="inpFileUrl" type="text" placeholder="上传后返回地址" />
            </div>
            <div class="col-sm-4">
                <input type="file" id="inpFileControl" name="file" />
            </div>
            <div class="col-sm-2">
                <input id="inpSubmit" name="inpSubmit" type="button" value="提交" />
            </div>
        </div>
    </fieldset>
</form>
 

Form表单提交Post到远程Api代码二中,这个很顺利。下边是Html

 
<form action="http://localhost:19420/Api/ReceiveFileTest" method="post" enctype="multipart/form-data">
    <fieldset>
        <legend>Form表单提交到远程服务器Api</legend>
        <div class="form-group">
            <label class="col-sm-2 control-label" for="ds_host">选择文件</label>
            <div class="col-sm-4">
                <input class="form-control" id="inpFileUrl2" name="inpFileUrl2" type="text" placeholder="上传后返回地址" />
            </div>
            <div class="col-sm-4">
                <input type="file" name="file" />
            </div>
            <div class="col-sm-2">
                <input type="submit" value="提交" />
            </div>
        </div>
    </fieldset>
</form>
 

Api服务部署时需要考虑到客户端提交file时有跨域问题,快捷方法是在IIS中设置Access-Control-Allow-Origin:*,但这样做有安全问题。

网站上传文件大小默认4M,所以网站配置文件中需要设置,如下:

  <system.web>
    <httpRuntime maxRequestLength="409600" />
  </system.web>

服务器IIS上传文件大小也有限制,也要做设置,如下:

 
  <system.webServer>
    <security>
      <requestFiltering>
        <!--2G/2072576000|500M/518144000-->
        <requestLimits maxAllowedContentLength="518144000"/>
      </requestFiltering>
    </security>
  </system.webServer>
 

以上是我近2天的研究成果,不算完善还需要改进。

Api中使用了三方框架RestSharp,请在解决方案中找NuGet添加。

外部操作类用到的方法有,一:

 
        /// <summary>
        /// 拼装JSON
        /// </summary>
        static public string GetResultJson(bool result, string msg, Object data)
        {
            string resultJson = string.Empty;
            Hashtable ht = new Hashtable();
            try
            {
                ht.Add("result", result);
                ht.Add("msg", msg);
                ht.Add("data", data);
            }
            catch (Exception ex)
            {
                ht.Add("result", false);
                ht.Add("msg", ex.Message);
            }
            resultJson = Newtonsoft.Json.JsonConvert.SerializeObject(ht);
            return resultJson;
        }
 

 注意:Html代码中file控件需要有name=“file”属性,否则提交后Api获取不到file对象,如下。

<input type="file" name="file" />

第三种方法是form表单提交调用mvc的control,control在调用Api代码二,这个测试失败,能上传文件,但上传的文件不能打开是损坏的,正在想办法解决。

下边是第三种方法的Html代码:

 
<form action="/Test/UploadFileTest" method="post" enctype="multipart/form-data">
    <fieldset>
        <legend>Form表单提交到本地Control</legend>
        <div class="form-group">
            <label class="col-sm-2 control-label" for="ds_host">选择文件</label>
            <div class="col-sm-4">
                <input class="form-control" id="inpFileUrl3" name="inpFileUrl3" type="text" placeholder="上传后返回地址" />
            </div>
            <div class="col-sm-4">
                <input type="file" name="file" />
            </div>
            <div class="col-sm-2">
                <input type="submit" value="提交" />
            </div>
        </div>
    </fieldset>
</form>
 

下边是第三种方法的Control代码:

 
       [HttpPost]
        public ActionResult UploadFileTest()
        {
            string result = string.Empty;

            string apiUrl = "http://localhost:19420/Api/ReceiveFileTest";
            string contentType = "application/octet-stream";

            var files = new List<string>();
            foreach (string f in Request.Files)
            {
                var file = Request.Files[f];
                var request = new RestRequest(Method.POST);
                request.AlwaysMultipartFormData = true;
                //request.AddParameter("fileName", file.FileName);
                Stream postStream = file.InputStream;
                byte[] b = new byte[postStream.Length];
                request.AddFile("file", b, file.FileName, contentType);

                var restClient = new RestClient { BaseUrl = new Uri(apiUrl) };
                string res = string.Empty;
                IRestResponse<Object> response = restClient.Execute<Object>(request);
                if (response.StatusCode == HttpStatusCode.OK)
                {
                    res = response.Content;
                }
                else
                {
                    res = string.Format("{0}:{1}", response.StatusCode, response.Content);
                }
                files.Add(res);
            }
            if (files.Count > 0) {
                result = string.Join(",", files.ToArray());
            }

            return Json(result);
        }
 

原文地址:https://www.cnblogs.com/zoujinhua/p/12692011.html