在一个应用程序中,我正在开发RESTful API,我们希望客户端以JSON形式发送数据。这个应用程序的一部分要求客户端上传一个文件(通常是一个图像)以及关于图像的信息。
我很难在一个请求中找到这种情况。是否可以将文件数据Base64转换为JSON字符串?我是否需要向服务器发送2次帖子?我不应该使用JSON吗?
顺便说一句,我们在后端使用Grails,这些服务是由本地移动客户端(iPhone、Android等)访问的,如果有什么不同的话。
在一个应用程序中,我正在开发RESTful API,我们希望客户端以JSON形式发送数据。这个应用程序的一部分要求客户端上传一个文件(通常是一个图像)以及关于图像的信息。
我很难在一个请求中找到这种情况。是否可以将文件数据Base64转换为JSON字符串?我是否需要向服务器发送2次帖子?我不应该使用JSON吗?
顺便说一句,我们在后端使用Grails,这些服务是由本地移动客户端(iPhone、Android等)访问的,如果有什么不同的话。
当前回答
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.
其他回答
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。
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.
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。
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());