what is the benefit of transform a string in this cod to Gson and then send it through socket in contrast without Gson?
String s ="abc";
DataOutputStream dos=new DataOutputStream();
Gson gson=new Gson();
String json= gson.toJson(s);
dos.writeUtf(json);
Related
I have a controller like so that accepts a MultipartFile and json object:
#PostMapping(value = "/v1/submit")
public ResponseEntity submit(
#RequestParam(value="myFile", required = true) MultipartFile myFile
, #Valid #RequestPart(value="fileMeta", required=true) FileMeta fileMeta
){
I need to forward this to a new url using an okhttpclient post with a Multipartbody containing both myFile and fileMeta objects:
OkHttpClient client = new OkHttpClient();
MultipartBody requestBody = new MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart("myFile", myFile.getName(), okhttp3.RequestBody.create(file, MediaType.parse("pdf"))
.addFormDataPart("fileMeta", fileMeta)
.build();
I am getting following error:
Cannot resolve method 'create(org.springframework.web.multipart.MultipartFile, okhttp3.MediaType)'
The method definition of OkHttp's RequestBody create is the following: create(MediaType contentType, byte[] content). It expects the first the MediaType and second the payload (either as byte[], File or other formats).
So you first have to switch the order of the method arguments and second convert the MultipartFile from Spring to a proper format that the create() method accepts, e.g. byte[] or File:
OkHttpClient client = new OkHttpClient();
MultipartBody requestBody = new MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart("myFile", myFile.getName(), RequestBody.create(MediaType.parse("pdf"), file)
.addFormDataPart("fileMeta", fileMeta)
.build();
There are already multiple solutions available on StackOverflow to convert MultipartFile to File: How to convert a multipart file to File?
UPDATE: Example for using RestTemplate
#RestController
public class FileSendingController {
#PostMapping("/files")
public void streamFile(#RequestParam("file") MultipartFile file) {
MultiValueMap<String, Object> body = new LinkedMultiValueMap<>();
body.add("file", file);
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
HttpEntity<MultiValueMap<String, Object>> requestEntity = new HttpEntity<>(body, headers);
RestTemplate restTemplate = new RestTemplate();
restTemplate.postForEntity("http://upload.to", requestEntity, String.class);
}
}
I need to send file to a POST service from the server side code.
The content of the file I need to send is in String format.
I don't want to create the file in the disk.
I can't find the way to send file without creating it in the disk.
I prefer not to create a TEMP file but this is what I managed to do.
How do I send file without saving it to disk, not even as TEMP file?
This is the code:
String fileContent = generateFile();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
headers.add("apikey","myapikey");
File tmpFile = File.createTempFile("test", ".tmp");
FileWriter writer = new FileWriter(tmpFile);
writer.write(fileContent);
writer.close();
BufferedReader reader = new BufferedReader(new FileReader(tmpFile));
reader.close();
FileSystemResource fsr = new FileSystemResource(tmpFile);
MultiValueMap<String, Object> body
= new LinkedMultiValueMap<>();
body.add("file",fsr);
HttpEntity<MultiValueMap<String, Object>> requestEntity
= new HttpEntity<>(body, headers);
String serverUrl = "https://api.com/api";
RestTemplate restTemplate = new RestTemplate();
ResponseEntity<String> response = restTemplate
.postForEntity(serverUrl, requestEntity, String.class);
return response.getBody();
POSTMAN screenshot that I use to test the API and it works perfect
Use a ByteArrayResource instead:
String fileContent = generateFile();
ByteArrayResource bar = new ByteArrayResource(fileContent.getBytes());
This way you will not have to create any resource on disk but keep it in memory instead.
I'm trying to POST a multipart/form-data using Spring RestTemplate with a byte array as the file to upload and it keeps failing (Server rejects with different kinds of errors).
I'm using a MultiValueMap with ByteArrayResource. Is there something I'm missing?
Yes there is something missing.
I have found this article:
https://medium.com/#voziv/posting-a-byte-array-instead-of-a-file-using-spring-s-resttemplate-56268b45140b
The author mentions that in order to POST a byte array using Spring RestTemplate one needs to override getFileName() of the ByteArrayResource.
Here is the code example from the article:
private static void uploadWordDocument(byte[] fileContents, final String filename) {
RestTemplate restTemplate = new RestTemplate();
String fooResourceUrl = "http://localhost:8080/spring-rest/foos"; // Dummy URL.
MultiValueMap<String, Object> map = new LinkedMultiValueMap<String, Object>();
map.add("name", filename);
map.add("filename", filename);
// Here we
ByteArrayResource contentsAsResource = new ByteArrayResource(fileContents) {
#Override
public String getFilename() {
return filename; // Filename has to be returned in order to be able to post.
}
};
map.add("file", contentsAsResource);
// Now you can send your file along.
String result = restTemplate.postForObject(fooResourceUrl, map, String.class);
// Proceed as normal with your results.
}
I tried it and it works!
I added an issue to send a request from java client to Python service in FastApi and sending a ByteArrayResource instaead of simple byte[] fixed the issue.
FastAPI server returned: "Expected UploadFile, received: <class 'str'>","type":"value_error""
I am using spring rest template to send json array as request. Source code to send request is as follow:
JSONArray jsonArray = new JSONArray();
for (Iterator iterator = itemlist.iterator(); iterator.hasNext();) {
Item item = (Item)iterator.next();
JSONObject formDetailsJson = new JSONObject();
formDetailsJson.put("id", item.getItemConfId());
formDetailsJson.put("name", item.getItems().getItemName());
formDetailsJson.put("price", item.getPrice());
formDetailsJson.put("Cost",item.getCost());
jsonArray.put(formDetailsJson);
}
List<MediaType> acceptableMediaTypes = new ArrayList<MediaType>();
acceptableMediaTypes.add(MediaType.APPLICATION_JSON);
// Prepare header
HttpHeaders headers = new HttpHeaders();
headers.setAccept(acceptableMediaTypes);
// Pass the new person and header
HttpEntity<JSONArray> entity = new HttpEntity<JSONArray>(jsonArray, headers);
System.out.println("Json Object : "+entity);
// Send the request as POST
try {
ResponseEntity<String> result = restTemplate.exchange("my url", HttpMethod.POST, entity, String.class);
} catch (Exception e) {
logger.error(e);
return "Connection not avilable please try again";
}
And to accept request:
#RequestMapping(value = "/testStock", method = RequestMethod.POST,headers="Accept=application/xml, application/json")
public #ResponseBody int testStock(#RequestBody List<ItemList> jsonArray) {
logger.debug("Received request to connect ms access : "+jsonArray.size());
//int returnSizecount = stockList.getStocklst().size();
return 1;
}
The problem is that it giving me following error:
Could not write request: no suitable HttpMessageConverter found for request type [org.json.JSONArray].Any suggestion is greatly acceptable.
There are no MessageConverter for JSONArray, so I suggest do the following.
HttpEntity<JSONArray> entity = new HttpEntity<JSONArray>(jsonArray, headers);
Convert Class JSONArray to String, and add that to HttpEntity, you know use toString
java.lang.String toString()
Make a JSON text of this JSONArray.
HttpEntity entity = new HttpEntity(jsonArray.toString(), headers);
Or change to Jackson implementation Spring have support to that. XD
If you dont want to do the above, consider create your own implementation of messageConverter, that will work but is harder
update
HttpHeaders headers = new HttpHeaders();
headers.setAccept(acceptableMediaTypes);
headers.setContentType(MediaType.APPLICATION_JSON);
update 2 Change endpoint to.
#RequestMapping(value = "/testStock", method = RequestMethod.POST)
public #ResponseBody int testStock(#RequestBody String jsonArray) {
you need to have httpmessageconverter configured for your resttemplate, please read my post for configuring http message conveter for you webservice
http://stackoverflow.com/questions/19963127/new-to-spring-and-jackson-2-what-does-this-bean-declaration-allow-for-in-a-spri/19973636#19973636.
and for you problem to convert your http request to json you might add this entry in your restemplate configuration
<bean id="jsonMessageConverter" class="org.springframework.http.converter.json.MappingJackson2HttpMessageConverter"/>
The error is quite straightforward. You do not have a converter for the JSONArray. Converting the array to a String (using toString) did help you here, but there is a better way:
Just add a converter for the json.org objects:
Add this to your pom.xml
<dependency>
<groupId>com.fasterxml.jackson.datatype</groupId>
<artifactId>jackson-datatype-json-org</artifactId>
</dependency>
And then on your ObjectMapper add the JsonOrgModule:
mapper.registerModule(new JsonOrgModule());
I am trying to post data using Spring RestTemplate as below:
MultiValueMap<String, String> parameters = new LinkedMultiValueMap<String, String>();
parameters.add("name1", "value1");
parameters.add("name2", "value2");
HttpMessageConverter<String> stringConverter = new StringHttpMessageConverter();
FormHttpMessageConverter formConverter = new FormHttpMessageConverter();
List<HttpMessageConverter<?>> msgConverters = new ArrayList<HttpMessageConverter<?>>();
msgConverters.add(formConverter);
msgConverters.add(stringConverter);
restTemplate.setMessageConverters(msgConverters);
String xml = restTemplate.postForObject(myurl, parameters, String.class);
On the server part, I am using a simple servlet to handle request as follow:
String name1 = request.getParameter("name1");
The server returns the xml as String.
When I used HashMap instead of MultiValueMap without Converter, the parameters are null on the server side. But after using the above code, I am getting error
Cannot extract response: no Content-Type found
Can you plz provide me a simple example to achieve what I want.
Here is what I used to format data for the Spring POST:
//FormHttpMessageConverter
is used to construct form parameters to POST on the URI
HttpMessageConverter<?> formHttpMessageConverter = new FormHttpMessageConverter();
HttpMessageConverter<?> stringHttpMessageConverter = new StringHttpMessageConverter();
List<HttpMessageConverter> msgConverters = new ArrayList<HttpMessageConverter>();
msgConverters.add(formHttpMessageConverter);
msgConverters.add(stringHttpMessageConverter);
// Prepare acceptable media type
List<MediaType> acceptableMediaTypes = new ArrayList<MediaType>();
acceptableMediaTypes.add(MediaType.ALL);
// Prepare header
HttpHeaders headers = new HttpHeaders();
headers.setAccept(acceptableMediaTypes);
HttpEntity<MultiValueMap<String,String>> httpEntity = new HttpEntity<MultiValueMap<String,String>>(map,headers);
ResponseEntity<String> resp = restTemplate.exchange("https://risk.XXXX.XXXXXX.net",HttpMethod.POST,httpEntity,String.class);