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""
Related
I am trying to fetch live data from NSE options trading. Below code is not working and the request made is stuck without any response.
Any workaround on this?
public void getLiveBankNiftyData() {
String RESOURCE_PATH = "https://www.nseindia.com/api/option-chain-indices?symbol=BANKNIFTY";
ResponseEntity<Object[]> responseEntity = restTemplate.getForEntity(RESOURCE_PATH, Object[].class);
Object[] objects = responseEntity.getBody();
}
i tried this
// request url
String url = "https://www.nseindia.com/api/option-chain-indices?symbol=BANKNIFTY";
// create an instance of RestTemplate
RestTemplate restTemplate = new RestTemplate();
// make an HTTP GET request
String json = restTemplate.getForObject(url, String.class);
// print json
System.out.println(json);
I found a way out. Instead of using RestTemplate I used WebClient and this solved the issue.
I'm trying to use Spring WebClient to make some basic REST API calls. I'm getting an error that the request is malformed, but I can't tell exactly why. Is there any way to easily log the contents of the request (really, just the request body)? Everything I find online is super complicated. Here's what I have:
LinkedMultiValueMap params = new LinkedMultiValueMap();
params.add("app_id", getOneSignalAppId());
params.add("included_segments", inSegment);
params.add("content_available", true);
params.add("contents", new LinkedMultiValueMap() {{
add("en", inTitle);
}});
BodyInserters.MultipartInserter inserter = BodyInserters.fromMultipartData(params);
WebClient client = WebClient.builder()
.baseUrl("https://onesignal.com")
.defaultHeader(HttpHeaders.AUTHORIZATION, "Basic " + getOneSignalKey())
.defaultHeader(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
.defaultHeader(HttpHeaders.ACCEPT, MediaType.APPLICATION_JSON_VALUE)
.build();
Mono<NotificationResponse> result = client
.post()
.uri("/api/v1/notifications")
.body(inserter)
.retrieve()
.bodyToMono(NotificationResponse.class);
I just want a string of the JSON that will be inserted into the request body.
You can create your own wrapper/proxy class around the JSON encoder (assuming you're using JSON) and intercept the serialized body before it is sent into the intertubes.
If your request is going to send JSON.
Specifically, you would extend the encodeValue method (or encodeValues in case of streaming data) of Jackson2JsonEncoder (the default encoder). Then you can do with that data what you wish, such as logging etc. And you could even do this conditionally based on environment/profile.
This custom logging-encoder can be specified when creating the WebClient, by providing it as a codec:
CustomBodyLoggingEncoder bodyLoggingEncoder = new CustomBodyLoggingEncoder();
WebClient.builder()
.codecs(clientDefaultCodecsConfigurer -> {
clientDefaultCodecsConfigurer.defaultCodecs().jackson2JsonEncoder(bodyLoggingEncoder);
clientDefaultCodecsConfigurer.defaultCodecs().jackson2JsonDecoder(new Jackson2JsonDecoder(new ObjectMapper(), MediaType.APPLICATION_JSON));
})
...
I made a blog post about this. You might be able to find the encoder for Multipart data and apply similar principles
For completeness, the encoder might look something like this:
public class CustomBodyLoggingEncoder extends Jackson2JsonEncoder {
#Override
public DataBuffer encodeValue(final Object value, final DataBufferFactory bufferFactory,
final ResolvableType valueType, #Nullable final MimeType mimeType, #Nullable final Map<String, Object> hints) {
// Encode/Serialize data to JSON
final DataBuffer data = super.encodeValue(value, bufferFactory, valueType, mimeType, hints);
// This is your code:
SomethingAmazing.doItWithThisData(extractBytes(data));
// Return the data as normal
return data;
}
private byte[] extractBytes(final DataBuffer data) {
final byte[] bytes = new byte[data.readableByteCount()];
data.read(bytes);
// We've copied the data above to our array, but must reset the buffer for actual usage
data.readPosition(0);
return bytes;
}
}
Hope that helps somehow!
I want to make a multipart request to some external API (created using Spring Boot) but all I get is Required request part 'file' is not present.
I know the source code of the external API but I can't modify it. It looks like this:
#PostMapping("/upload")
public ResponseEntity handleFileUpload(#RequestParam("file") MultipartFile file){
return ResponseEntity.ok().build();
}
And from my application I create and send requests exactly like on the following snippet:
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
MultiValueMap<String, Object> body
= new LinkedMultiValueMap<>();
body.add("file", "dupa".getBytes());
HttpEntity<MultiValueMap<String, Object>> requestEntity
= new HttpEntity<>(body, headers);
RestTemplate restTemplate = new RestTemplate();
ResponseEntity<String> response = restTemplate
.postForEntity("http://api:8080/upload", requestEntity, String.class);
return response.getBody();
What's the reason it doesn't work? The above code rewritten using Apache HttpClient works like charm.
You basically have two options, the solution with byte array:
map.add("file", new ByteArrayResource(byteArrayContent) {
#Override
public String getFilename() {
return "yourFilename";
}
});
I remember having a problem with just adding a byte array, so you need to have a filename too and use ByteArrayResource.
Or adding a File:
map.add("file", new FileSystemResource(file));
I really like the solution I have with RestTemplate but soon it will be depreciated with future spring releases. I am trying to send some text to a third party api using WebClient
String text = URLEncoder.encode(text,"UTF-8");
WebClient webClient = WebClient.builder()
.baseUrl(BASE_URL)
.defaultHeader("Key","af999-e99-4456-b556-4ef9947383d")
.defaultHeader("src", srcLang)
.defaultHeader("tgt", tgtLang)
.defaultHeader("text", text)
.build();
Then send a post here:
Mono<String> response = webClient.post().uri("/google/rtv/text")
.retrieve()
.bodyToMono(String.class);
Trying to parse based off of the legacy response:
private String parseJson( Mono<String> response) {
ObjectMapper mapper = new ObjectMapper();
JsonNode root = null;
JsonNode review = null;
//TODO: create an object and map it here. We need to save the original review too.
try {
root = mapper.readTree(response.toString());
review = root.path("message");
} catch (IOException e) {
e.printStackTrace();
}
return review.asText();
}
Later I need to parse the response but right now I am getting an error saying:
com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'MonoFlatMap': was expecting ('true', 'false' or 'null')
at [Source: (String)"MonoFlatMap"; line: 1, column: 23]
and later:
java.lang.NullPointerException: null
What I am trying to accomplish is something like I have done with RestTemplate.
Like so:
UriComponentsBuilder builder = UriComponentsBuilder
.fromUriString(URL)
.queryParam("src", src)
.queryParam("tgt", tgt)
.queryParam("text", text);
ResponseEntity<String> response = restTemplate.exchange(builder.toUriString(), HttpMethod.GET, request, String.class);
then set my header for the subscription globally.
private ClientHttpResponse intercept(HttpRequest request, byte[] body,
ClientHttpRequestExecution execution) throws IOException {
request.getHeaders().add("Key","af999-e99-4456-b556-4ef9947383d");
ClientHttpResponse response = execution.execute(request, body);
return response;
}
#Bean
public RestTemplate restTemplate() {
RestTemplate restTemplate = new RestTemplate();
restTemplate.setInterceptors(Collections.singletonList(this::intercept));
return restTemplate;
}
Advice?
The problem happens here:
root = mapper.readTree(response.toString());
This code snippet is trying to serialize a Mono<String> as a String, when a Mono is a reactive type that can provide that String value eventually.
You could call response.block() and getting the resulting String, but this would be a blocking call and Reactor forbids that if in the middle of a reactive execution. This is done for good reasons, since this will block one of the few threads that your web application is using and can cause it to stop serving other requests.
You could instead have something like:
Mono<String> review = response.map(r -> parseJson(r);
And then reuse that new value down the line.
Note that WebClient natively supports JSON deserialization and you could deserialize the whole payload like so:
Mono<Review> review = webClient.post().uri("/google/rtv/text")
.retrieve()
.bodyToMono(Review.class);
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());