这是我的方法API(我使用示例)-如你所见，你我没有在API中使用任何file_id(上传到服务器的文件标识符):

在服务器上创建照片对象: 职位:/项目/ {project_id} /照片 正文:{名称:"some_schema.jpg"，评论:"blah"} 回应:photo_id 上传文件(注意文件是单数形式，因为每张照片只有一个): 职位:/项目/ {project_id} / {photo_id} /文件/照片 正文:要上传的文件 响应:

举个例子:

阅读照片列表 得到:/项目/ {project_id} /照片 回复:[照片，照片，照片，…](对象数组) 阅读一些照片细节 得到:/项目/ {project_id} /照片/ {photo_id} 响应:{id: 666，名称:' some_schema.jpg'，评论:'blah'}(照片对象) 读取照片文件 得到:/项目/ {project_id} / {photo_id} /文件/照片 响应:文件内容

So the conclusion is that, first you create an object (photo) by POST, and then you send second request with the file (again POST). To not have problems with CACHE in this approach we assume that we can only delete old photos and add new - no update binary photo files (because new binary file is in fact... NEW photo). However if you need to be able to update binary files and cache them, then in point 4 return also fileId and change 5 to GET: /projects/{project_id}/photos/{photo_id}/files/{fileId}.

请确保您有以下导入。当然还有其他标准的进口

import org.springframework.core.io.FileSystemResource


    void uploadzipFiles(String token) {

        RestBuilder rest = new RestBuilder(connectTimeout:10000, readTimeout:20000)

        def zipFile = new File("testdata.zip")
        def Id = "001G00000"
        MultiValueMap<String, String> form = new LinkedMultiValueMap<String, String>()
        form.add("id", id)
        form.add('file',new FileSystemResource(zipFile))
        def urld ='''http://URL''';
        def resp = rest.post(urld) {
            header('X-Auth-Token', clientSecret)
            contentType "multipart/form-data"
            body(form)
        }
        println "resp::"+resp
        println "resp::"+resp.text
        println "resp::"+resp.headers
        println "resp::"+resp.body
        println "resp::"+resp.status
    }

FormData对象:使用Ajax上传文件

XMLHttpRequest Level 2增加了对新的FormData接口的支持。 FormData对象提供了一种方法，可以轻松地构造一组表示表单字段及其值的键/值对，然后可以使用XMLHttpRequest send()方法轻松地发送这些字段。

function AjaxFileUpload() {
    var file = document.getElementById("files");
    //var file = fileInput;
    var fd = new FormData();
    fd.append("imageFileData", file);
    var xhr = new XMLHttpRequest();
    xhr.open("POST", '/ws/fileUpload.do');
    xhr.onreadystatechange = function () {
        if (xhr.readyState == 4) {
             alert('success');
        }
        else if (uploadResult == 'success')
             alert('error');
    };
    xhr.send(fd);
}

https://developer.mozilla.org/en-US/docs/Web/API/FormData

由于唯一缺少的例子是ANDROID的例子，我将添加它。 该技术使用一个自定义AsyncTask，应该在Activity类中声明。

private class UploadFile extends AsyncTask<Void, Integer, String> {
    @Override
    protected void onPreExecute() {
        // set a status bar or show a dialog to the user here
        super.onPreExecute();
    }

    @Override
    protected void onProgressUpdate(Integer... progress) {
        // progress[0] is the current status (e.g. 10%)
        // here you can update the user interface with the current status
    }

    @Override
    protected String doInBackground(Void... params) {
        return uploadFile();
    }

    private String uploadFile() {

        String responseString = null;
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("http://example.com/upload-file");

        try {
            AndroidMultiPartEntity ampEntity = new AndroidMultiPartEntity(
                new ProgressListener() {
                    @Override
                        public void transferred(long num) {
                            // this trigger the progressUpdate event
                            publishProgress((int) ((num / (float) totalSize) * 100));
                        }
            });

            File myFile = new File("/my/image/path/example.jpg");

            ampEntity.addPart("fileFieldName", new FileBody(myFile));

            totalSize = ampEntity.getContentLength();
            httpPost.setEntity(ampEntity);

            // Making server call
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();

            int statusCode = httpResponse.getStatusLine().getStatusCode();
            if (statusCode == 200) {
                responseString = EntityUtils.toString(httpEntity);
            } else {
                responseString = "Error, http status: "
                        + statusCode;
            }

        } catch (Exception e) {
            responseString = e.getMessage();
        }
        return responseString;
    }

    @Override
    protected void onPostExecute(String result) {
        // if you want update the user interface with upload result
        super.onPostExecute(result);
    }

}

所以，当你想上传文件时，只需调用:

new UploadFile().execute();

我想发送一些字符串到后端服务器。我没有使用json与multipart，我已经使用请求参数。

@RequestMapping(value = "/upload", method = RequestMethod.POST)
public void uploadFile(HttpServletRequest request,
        HttpServletResponse response, @RequestParam("uuid") String uuid,
        @RequestParam("type") DocType type,
        @RequestParam("file") MultipartFile uploadfile)

Url看起来像这样

http://localhost:8080/file/upload?uuid=46f073d0&type=PASSPORT

我在文件上传时传递了两个参数(uuid和type)。 希望这将帮助谁没有复杂的json数据发送。

我在这里问了一个类似的问题:

如何使用REST web服务上传带有元数据的文件?

你基本上有三个选择:

Base64 encode the file, at the expense of increasing the data size by around 33%, and add processing overhead in both the server and the client for encoding/decoding. Send the file first in a multipart/form-data POST, and return an ID to the client. The client then sends the metadata with the ID, and the server re-associates the file and the metadata. Send the metadata first, and return an ID to the client. The client then sends the file with the ID, and the server re-associates the file and the metadata.