I am trying to write Resteasy Client Class to make simple POST multipart/form-data request to upload file. So POST request must contain Content-Disposition: form-data; name="files"; filename="myfile.txt" parameters.
I am able to set this statically with #PartFilename annotation for field in entity (value) class. like this
public class UploadStreamMultipartBody {
#FormParam("files")
#PartFilename(value = "myfile.txt")
#PartType(MediaType.APPLICATION_OCTET_STREAM)
public InputStream file;
}
But I cannot realize how to make it dynamically so I can provide file name in RUNTIME 🤔
Cause annotation value is read once in compile time.
Use MultipartFormDataOutput instead.
Interface example:
#POST
#Path("/document-store")
#Consumes(MediaType.MULTIPART_FORM_DATA)
void uploadDocument(#MultipartForm MultipartFormDataOutput data);
Usage:
var form = new MultipartFormDataOutput();
form.addFormData(/*key=*/"file", inputStream, new MediaType("application", "pdf"), fileName);
documentStoreClient.uploadDocument(form);
I am trying to make an Http Request using RestTemplate, and it keeps on giving me the error: 'HttpHeaders' has private access in 'org.apache.http.HttpHeaders'
I am simply trying to write this line:
HttpHeaders headers = new HttpHeaders();
The package name is wrong, in order to add headers when using Spring restTemplate, you should use org.springframework.http.HttpHeaders.HttpHeaders instead of org.apache.http.HttpHeaders.
The following is the code snippet that adds request headers.
// request resource
HttpHeaders headers = new HttpHeaders();
headers.set("headerName", "headerValue");
HttpEntity entity = new HttpEntity(headers);
ResponseEntity<String> response = restTemplate.exchange("https://example.com", HttpMethod.GET, entity, String.class);
The constructor in org.apache.http.HttpHeaders is a private constructor - see source code clone {here}. Since you are trying to invoke a private attribute, that error message is expected.
Attaching relevant code snippet for posterity:
public final class HttpHeaders {
private HttpHeaders() {
}
// ....
// bunch of defined constants
// ....
}
The rationale behind this class is specified in the class docstring,
/**
* Constants enumerating the HTTP headers. All headers defined in RFC1945 (HTTP/1.0), RFC2616 (HTTP/1.1), and RFC2518
* (WebDAV) are listed.
*
* #since 4.1
*/
which is not the what you are trying to achieve here. If you wish to make a remote request, using apache library, with a request that contains headers, please follow {this example}. Adding relevant code snippet for posterity:
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet(SAMPLE_URL);
request.setHeader(HttpHeaders.CONTENT_TYPE, "application/json");
client.execute(request);
If you are using >=4.3 of HttpClient, you would want to do something like this:
HttpUriRequest request = RequestBuilder.get()
.setUri(SAMPLE_URL)
.setHeader(HttpHeaders.CONTENT_TYPE, "application/json")
.build();
To remove compilation errors with
HttpHeaders headers = new HttpHeaders();
make sure you haven't imported HttpHeaders from apache http clients.
import org.apache.http.HttpHeaders;
this will not work as HttpHeaders has a private constructor.
Your import statement should be this:
import org.springframework.http.HttpHeaders;
And then you can add headers using the add(String headerName, String headerValue) method.
Source: Link
I am trying to consume the following HTTPS endpoints from Yahoo Weather Service:
Yahoo Weather Service API
I am doing some special query according to the API to get the current weather at some parametrized location.
#Service("weatherConditionService")
public class WeatherConditionServiceImpl implements WeatherConditionService {
private static final String URL = "http://query.yahooapis.com/v1/public/yql";
public WeatherCondition getCurrentWeatherConditionsFor(Location location) {
RestTemplate restTemplate = new RestTemplate();
StringBuilder stringBuilder = new StringBuilder();
stringBuilder.append(URL);
stringBuilder.append("?q=select%20item.condition%20from%20weather.forecast%20where%20woeid%20in%20(select%20woeid%20from%20geo.places(1)%20where%20text%3D%22");
// TODO: Validate YQL query injection
stringBuilder.append(location.getName());
stringBuilder.append("%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys");
WeatherQuery weatherQuery = restTemplate.getForObject(stringBuilder.toString(), WeatherQuery.class);
// TODO: Test Json mapping response
Condition condition = weatherQuery.getQuery().getResults().getChannel().getItem().getCondition();
return new WeatherCondition(condition.getDate(), Integer.parseInt(condition.getTemp()), condition.getText());
}
Location is a class that provides the attribute "name" that is a String description of the location, such as "New York" or "Manila".
Condition an other classes just map the returning object.
When executing I get the following HTTP response:
org.springframework.web.client.HttpClientErrorException: 403 Forbidden
So this means I am not authorized to access the resource from what I understand.
The URL works great if I just copy & paste it in a web browser:
Yahoo Weather Query
I think that mapping is not a problem since I am not getting "400" (Bad Request) but "403" (Forbidden)
There must be some error on the way I use the RestTemplate object. I am researching but I can't find an answer.
The docs say you need an api key. But when I make a call like this:
fetch('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20weather.forecast%20where%20woeid%20in%20(select%20woeid%20from%20geo.places(1)%20where%20text%3D%22nome%2C%20ak%22)&format=json&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys')
.then(resp=> resp.json())
.then((res)=>console.log(res.query.results))
https://repl.it/NeoM
It works fine without one. Perhaps you've been blackisted for hitting the api too often.
Your code seems fine.
I finally found the answer. It finally WAS a Bad Request because I needed to pass the parameters differently (not as part of the URL).
I found the answer here. Here goes the code for my particular Yahoo Weather API call return a String (I still will have to do some work to use the mapping).
private static final String URL = "http://query.yahooapis.com/v1/public/yql";
public String callYahooWeatherApi() {
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(URL)
.queryParam("q", "select wind from weather.forecast where woeid=2460286")
.queryParam("format", "json");
HttpEntity<?> entity = new HttpEntity<>(headers);
HttpEntity<String> response = restTemplate.exchange(
builder.build().encode().toUri(),
HttpMethod.GET,
entity,
String.class);
return response.getBody();
}
I know the question is weird. Unfortunately I have a service that requires everything to have the header ContentType=application/x-www-form-urlencoded, eventhough the body is JSON
I am trying to use JAX-RS 2.0 ClientBuilder to call it:
String baseUrl = "http://api.example.com/";
JSONObject body = new JSONObject();
body.put("key", "value");
Client client = ClientBuilder.newClient();
client.register(new LoggingFilter());
Builder builder = client.target(baseUrl).path("something").request();
Invocation inv = builder
.header("Content-type", MediaType.APPLICATION_FORM_URLENCODED)
.buildPost(Entity.json(body));
Response response = inv.invoke();
int status = response.getStatus();
// I get 415, unsupported media type (in this case is unexpected)
I have checked my logs and I eventhough I am setting application/x-www-form-urlencoded (via the MediaType) the request appearantly has the Content-type of application/json
How can I force the request to have the Content-type I want?
BTW: This is my custom logger:
public class LoggingFilter implements ClientRequestFilter {
private static final Logger LOG = Logger.getLogger(LoggingFilter.class.getName());
#Override
public void filter(ClientRequestContext requestContext) throws IOException {
LOG.log(Level.INFO, "body");
LOG.log(Level.INFO, requestContext.getEntity().toString());
LOG.log(Level.INFO, "headers");
LOG.log(Level.INFO, requestContext.getHeaders().toString());
}
}
And these are the logs I get:
com.acme.LoggingFilter I body
com.acme.LoggingFilter I {"key":"value"}
com.acme.LoggingFilter I headers
com.acme.LoggingFilter I {Content-type=[application/json]}
The problem with trying to use one of the static Entity helper methods is that it overrides any previous Content-Type header you may have set. In your current case, Entity.json automatically sets the header to application/json.
Instead of using the .json method, you can just use the general purpose Entity.entity(Object, MediaType) method. With your current case though, you can just do Entity.entity(body, MediaType.APPLICATION_FORM_URLENCODED_TYPE) though. The reason is that the client will look for a provider that knows how to serialize a JSONObject to application/x-www-form-urlencoded data, which there is none. So you will need to first serialize it to a String. That way the provider that handles application/x-www-form-urlencoded doesn't need to serialize anything. So just do
Entity.entity(body.toString(), MediaType.APPLICATION_FORM_URLENCODED_TYPE);
I want to create an employee information in the system by uploading an image along with employee data. I am able to do it with different rest calls using jersey. But I want to achieve in one rest call.
I provide below the structure. Please help me how to do in this regard.
#POST
#Path("/upload2")
#Consumes({MediaType.MULTIPART_FORM_DATA,MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON})
public Response uploadFileWithData(
#FormDataParam("file") InputStream fileInputStream,
#FormDataParam("file") FormDataContentDisposition contentDispositionHeader,
Employee emp) {
//..... business login
}
Whenever I am trying to do, I get error in Chrome postman. The simple structure of my Employee json is given below.
{
"Name": "John",
"Age": 23,
"Email": "john#gmail.com",
"Adrs": {
"DoorNo": "12-A",
"Street": "Street-11",
"City": "Bangalore",
"Country": "Karnataka"
}
}
However I can do it by making two different call, but I want to achieve in one rest call so that I can receive the file as well as the actual data of the employee.
Request you to help in this regard.
You can't have two Content-Types (well technically that's what we're doing below, but they are separated with each part of the multipart, but the main type is multipart). That's basically what you are expecting with your method. You are expecting mutlipart and json together as the main media type. The Employee data needs to be part of the multipart. So you can add a #FormDataParam("emp") for the Employee.
#FormDataParam("emp") Employee emp) { ...
Here's the class I used for testing
#Path("/multipart")
public class MultipartResource {
#POST
#Path("/upload2")
#Consumes({MediaType.MULTIPART_FORM_DATA})
public Response uploadFileWithData(
#FormDataParam("file") InputStream fileInputStream,
#FormDataParam("file") FormDataContentDisposition cdh,
#FormDataParam("emp") Employee emp) throws Exception{
Image img = ImageIO.read(fileInputStream);
JOptionPane.showMessageDialog(null, new JLabel(new ImageIcon(img)));
System.out.println(cdh.getName());
System.out.println(emp);
return Response.ok("Cool Tools!").build();
}
}
First I just tested with the client API to make sure it works
#Test
public void testGetIt() throws Exception {
final Client client = ClientBuilder.newBuilder()
.register(MultiPartFeature.class)
.build();
WebTarget t = client.target(Main.BASE_URI).path("multipart").path("upload2");
FileDataBodyPart filePart = new FileDataBodyPart("file",
new File("stackoverflow.png"));
// UPDATE: just tested again, and the below code is not needed.
// It's redundant. Using the FileDataBodyPart already sets the
// Content-Disposition information
filePart.setContentDisposition(
FormDataContentDisposition.name("file")
.fileName("stackoverflow.png").build());
String empPartJson
= "{"
+ " \"id\": 1234,"
+ " \"name\": \"Peeskillet\""
+ "}";
MultiPart multipartEntity = new FormDataMultiPart()
.field("emp", empPartJson, MediaType.APPLICATION_JSON_TYPE)
.bodyPart(filePart);
Response response = t.request().post(
Entity.entity(multipartEntity, multipartEntity.getMediaType()));
System.out.println(response.getStatus());
System.out.println(response.readEntity(String.class));
response.close();
}
I just created a simple Employee class with an id and name field for testing. This works perfectly fine. It shows the image, prints the content disposition, and prints the Employee object.
I'm not too familiar with Postman, so I saved that testing for last :-)
It appears to work fine also, as you can see the response "Cool Tools". But if we look at the printed Employee data, we'll see that it's null. Which is weird because with the client API it worked fine.
If we look at the Preview window, we'll see the problem
There's no Content-Type header for the emp body part. You can see in the client API I explicitly set it
MultiPart multipartEntity = new FormDataMultiPart()
.field("emp", empPartJson, MediaType.APPLICATION_JSON_TYPE)
.bodyPart(filePart);
So I guess this is really only part of a full answer. Like I said, I am not familiar with Postman So I don't know how to set Content-Types for individual body parts. The image/png for the image was automatically set for me for the image part (I guess it was just determined by the file extension). If you can figure this out, then the problem should be solved. Please, if you find out how to do this, post it as an answer.
See UPDATE below for solution
And just for completeness...
See here for more about MultiPart with Jersey.
Basic configurations:
Dependency:
<dependency>
<groupId>org.glassfish.jersey.media</groupId>
<artifactId>jersey-media-multipart</artifactId>
<version>${jersey2.version}</version>
</dependency>
Client config:
final Client client = ClientBuilder.newBuilder()
.register(MultiPartFeature.class)
.build();
Server config:
// Create JAX-RS application.
final Application application = new ResourceConfig()
.packages("org.glassfish.jersey.examples.multipart")
.register(MultiPartFeature.class);
If you're having problems with the server configuration, one of the following posts might help
What exactly is the ResourceConfig class in Jersey 2?
152 MULTIPART_FORM_DATA: No injection source found for a parameter of type public javax.ws.rs.core.Response
UPDATE
So as you can see from the Postman client, some clients are unable to set individual parts' Content-Type, this includes the browser, in regards to it's default capabilities when using FormData (js).
We can't expect the client to find away around this, so what we can do, is when receiving the data, explicitly set the Content-Type before deserializing. For example
#POST
#Path("upload2")
#Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFileAndJSON(#FormDataParam("emp") FormDataBodyPart jsonPart,
#FormDataParam("file") FormDataBodyPart bodyPart) {
jsonPart.setMediaType(MediaType.APPLICATION_JSON_TYPE);
Employee emp = jsonPart.getValueAs(Employee.class);
}
It's a little extra work to get the POJO, but it is a better solution than forcing the client to try and find it's own solution.
Another option is to use a String parameter and use whatever JSON library you use to deserialze the String to the POJO (like Jackson ObjectMapper). With the previous option, we just let Jersey handle the deserialization, and it will use the same JSON library it uses for all the other JSON endpoints (which might be preferred).
Asides
There is a conversation in these comments that you may be interested in if you are using a different Connector than the default HttpUrlConnection.
You can access the Image File and data from a form using MULTIPART FORM DATA By using the below code.
#POST
#Path("/UpdateProfile")
#Consumes(value={MediaType.APPLICATION_JSON,MediaType.MULTIPART_FORM_DATA})
#Produces(value={MediaType.APPLICATION_JSON,MediaType.APPLICATION_XML})
public Response updateProfile(
#FormDataParam("file") InputStream fileInputStream,
#FormDataParam("file") FormDataContentDisposition contentDispositionHeader,
#FormDataParam("ProfileInfo") String ProfileInfo,
#FormDataParam("registrationId") String registrationId) {
String filePath= "/filepath/"+contentDispositionHeader.getFileName();
OutputStream outputStream = null;
try {
int read = 0;
byte[] bytes = new byte[1024];
outputStream = new FileOutputStream(new File(filePath));
while ((read = fileInputStream.read(bytes)) != -1) {
outputStream.write(bytes, 0, read);
}
outputStream.flush();
outputStream.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
if (outputStream != null) {
try {
outputStream.close();
} catch(Exception ex) {}
}
}
}
When I tried #PaulSamsotha's solution with Jersey client 2.21.1, there was 400 error. It worked when I added following in my client code:
MediaType contentType = MediaType.MULTIPART_FORM_DATA_TYPE;
contentType = Boundary.addBoundary(contentType);
Response response = t.request()
.post(Entity.entity(multipartEntity, contentType));
instead of hardcoded MediaType.MULTIPART_FORM_DATA in POST request call.
The reason this is needed is because when you use a different Connector (like Apache) for the Jersey Client, it is unable to alter outbound headers, which is required to add a boundary to the Content-Type. This limitation is explained in the Jersey Client docs. So if you want to use a different Connector, then you need to manually create the boundary.
Your ApplicationConfig should register the MultiPartFeature.class from the glassfish.jersey.media.. so as to enable file upload
#javax.ws.rs.ApplicationPath(ResourcePath.API_ROOT)
public class ApplicationConfig extends ResourceConfig {
public ApplicationConfig() {
//register the necessary headers files needed from client
register(CORSConfigurationFilter.class);
//The jackson feature and provider is used for object serialization
//between client and server objects in to a json
register(JacksonFeature.class);
register(JacksonProvider.class);
//Glassfish multipart file uploader feature
register(MultiPartFeature.class);
//inject and registered all resources class using the package
//not to be tempered with
packages("com.flexisaf.safhrms.client.resources");
register(RESTRequestFilter.class);
}
I used file upload example from,
http://www.mkyong.com/webservices/jax-rs/file-upload-example-in-jersey/
in my resource class i have below method
#POST
#Path("/upload")
#Consumes(MediaType.MULTIPART_FORM_DATA)
public Response attachupload(#FormDataParam("file") byte[] is,
#FormDataParam("file") FormDataContentDisposition fileDetail,
#FormDataParam("fileName") String flename){
attachService.saveAttachment(flename,is);
}
in my attachService.java i have below method
public void saveAttachment(String flename, byte[] is) {
// TODO Auto-generated method stub
attachmentDao.saveAttachment(flename,is);
}
in Dao i have
attach.setData(is);
attach.setFileName(flename);
in my HBM mapping is like
<property name="data" type="binary" >
<column name="data" />
</property>
This working for all type of files like .PDF,.TXT, .PNG etc.,
The request type is multipart/form-data and what you are sending is essentially form fields that go out as bytes with content boundaries separating different form fields.To send an object representation as form field (string), you can send a serialized form from the client that you can then deserialize on the server.
After all no programming environment object is actually ever traveling on the wire. The programming environment on both side are just doing automatic serialization and deserialization that you can also do. That is the cleanest and programming environment quirks free way to do it.
As an example, here is a javascript client posting to a Jersey example service,
submitFile(){
let data = new FormData();
let account = {
"name": "test account",
"location": "Bangalore"
}
data.append('file', this.file);
data.append("accountKey", "44c85e59-afed-4fb2-884d-b3d85b051c44");
data.append("device", "test001");
data.append("account", JSON.stringify(account));
let url = "http://localhost:9090/sensordb/test/file/multipart/upload";
let config = {
headers: {
'Content-Type': 'multipart/form-data'
}
}
axios.post(url, data, config).then(function(data){
console.log('SUCCESS!!');
console.log(data.data);
}).catch(function(){
console.log('FAILURE!!');
});
},
Here the client is sending a file, 2 form fields (strings) and an account object that has been stringified for transport. here is how the form fields look on the wire,
On the server, you can just deserialize the form fields the way you see fit. To finish this trivial example,
#POST
#Path("/file/multipart/upload")
#Consumes({MediaType.MULTIPART_FORM_DATA})
public Response uploadMultiPart(#Context ContainerRequestContext requestContext,
#FormDataParam("file") InputStream fileInputStream,
#FormDataParam("file") FormDataContentDisposition cdh,
#FormDataParam("accountKey") String accountKey,
#FormDataParam("account") String json) {
System.out.println(cdh.getFileName());
System.out.println(cdh.getName());
System.out.println(accountKey);
try {
Account account = Account.deserialize(json);
System.out.println(account.getLocation());
System.out.println(account.getName());
} catch (Exception e) {
e.printStackTrace();
}
return Response.ok().build();
}