我最近切换到OkHttp
。切换后,下面的代码执行上传。
RequestBody requestBody = new MultipartBuilder()
.type(MultipartBuilder.FORM)
.addPart(
Headers.of("Content-Disposition", "form-data; name=\"qqfile\""),
RequestBody.create(
MediaType.parse(filename),
new File(filename)))
.build();
如果您比较图片,则第二张图片有multipartFiles
size = 0
。它应该是size = 1
。如何使用multipartHttpRequest
正确填充OkHttp
以使服务器接受成功上传?
控制器代码
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.springframework.http.MediaType;
import org.springframework.web.multipart.MultipartFile;
import org.springframework.web.multipart.MultipartHttpServletRequest;
import org.springframework.web.util.WebUtils;
@RequestMapping (
method = RequestMethod.POST,
value = "/upload",
produces = MediaType.APPLICATION_JSON_VALUE + ";charset=UTF-8"
)
public String upload(
HttpServletRequest request,
HttpServletResponse response
) throws IOException {
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
if (isMultipart) {
MultipartHttpServletRequest multipartHttpRequest =
WebUtils.getNativeRequest(request, MultipartHttpServletRequest.class);
final List<MultipartFile> files = multipartHttpRequest.getFiles("qqfile");
if (files.isEmpty()) {
LOG.error("qqfile name missing in request or no file uploaded");
return some error code here
}
MultipartFile multipartFile = files.iterator().next();
//process file code below
}
return failure;
}
答案 0 :(得分:3)
您可以更轻松地使用MultipartFile:
@RequestMapping(value = "/upload", method = RequestMethod.POST)
public String upload(@RequestParam("qqfile") MultipartFile file) throws IOException {
if (!file.isEmpty()) {
// ...
}
return "failure";
}
然后,用OkHttp:
RequestBody body = new MultipartBuilder()
.addFormDataPart("qqfile", filename, RequestBody.create(MediaType.parse("media/type"), new File(filename)))
.type(MultipartBuilder.FORM)
.build();
Request request = new Request.Builder()
.url("/path/to/your/upload")
.post(body)
.build();
OkHttpClient client = new OkHttpClient();
Response response = client.newCall(request).execute();
这对我很好。
小心使用MediaType.parse(filename),你必须传递一个有效的类型,如text / plain,application / json,application / xml ......
答案 1 :(得分:2)
Builder requestBodyBuilder = new MultipartBody.Builder()
.setType(MultipartBody.FORM);
File file= new File(FILE_PATH + FILE_NAME);
requestBodyBuilder.addFormDataPart("file", FILE_NAME, RequestBody.create(MultipartBody.FORM, file));
<块引用>
fileVO.getOriginalFlnm()
您可以省略此字段。
并且您还必须在标头中设置“MultipartHttpServletRequest”参数和消费、生产
@PostMapping(path = "/save", consumes = "multipart/*", produces = "application/json;charset=utf-8")
public boolean CONTROLLER(MultipartHttpServletRequest request, @RequestParam Map<String, Object> param) {
boolean result = SERVICE.save(request, param);
return result;
}