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.

我知道这个问题很老了，但在过去的几天里，我搜索了整个网络来解决这个问题。我有grails REST web服务和iPhone客户端发送图片，标题和描述。

我不知道我的方法是不是最好的，但是很简单。

我使用UIImagePickerController拍了一张照片，并使用请求的头标签将NSData发送给服务器，以发送图片的数据。

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"myServerAddress"]];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:UIImageJPEGRepresentation(picture, 0.5)];
[request setValue:@"image/jpeg" forHTTPHeaderField:@"Content-Type"];
[request setValue:@"myPhotoTitle" forHTTPHeaderField:@"Photo-Title"];
[request setValue:@"myPhotoDescription" forHTTPHeaderField:@"Photo-Description"];

NSURLResponse *response;

NSError *error;

[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

在服务器端，我使用以下代码接收照片:

InputStream is = request.inputStream

def receivedPhotoFile = (IOUtils.toByteArray(is))

def photo = new Photo()
photo.photoFile = receivedPhotoFile //photoFile is a transient attribute
photo.title = request.getHeader("Photo-Title")
photo.description = request.getHeader("Photo-Description")
photo.imageURL = "temp"    

if (photo.save()) {    

    File saveLocation = grailsAttributes.getApplicationContext().getResource(File.separator + "images").getFile()
    saveLocation.mkdirs()

    File tempFile = File.createTempFile("photo", ".jpg", saveLocation)

    photo.imageURL = saveLocation.getName() + "/" + tempFile.getName()

    tempFile.append(photo.photoFile);

} else {

    println("Error")

}

我不知道将来是否会有问题，但现在在生产环境中工作得很好。

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

您可以尝试使用https://square.github.io/okhttp/ library。 你可以设置请求主体为multipart，然后分别添加文件和json对象，如下所示:

MultipartBody requestBody = new MultipartBody.Builder()
                .setType(MultipartBody.FORM)
                .addFormDataPart("uploadFile", uploadFile.getName(), okhttp3.RequestBody.create(uploadFile, MediaType.parse("image/png")))
                .addFormDataPart("file metadata", json)
                .build();

        Request request = new Request.Builder()
                .url("https://uploadurl.com/uploadFile")
                .post(requestBody)
                .build();

        try (Response response = client.newCall(request).execute()) {
            if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);

            logger.info(response.body().string());

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}.

@RequestMapping(value = "/uploadImageJson", method = RequestMethod.POST)
    public @ResponseBody Object jsongStrImage(@RequestParam(value="image") MultipartFile image, @RequestParam String jsonStr) {
-- use  com.fasterxml.jackson.databind.ObjectMapper convert Json String to Object
}