Issue with creating resource by REST WS using multipart content-type

[mateusz.moska]

Hi,

I've got a problem with creating the simple resource by putting resource descriptor into multipart request. I have been looking for the solution on jasper server community, but I couldn't have found an answear. If I try to create a resource, the same as is in the documentation, I will get Error 400 Bad Request: "Invalid Resource Descriptor" and in the jasperserver.log file I've got an exception:

2012-12-10 11:41:55,054 ERROR RESTResource,http-18080-6:237 - Unexpected error during resource descriptor marshaling: Premature end of file.
org.xml.sax.SAXParseException: Premature end of file. at
org.apache.xerces.parsers.DOMParser.parse(Unknown Source) at org.apache.xerces.jaxp.DocumentBuilderImpl.parse(Unknown Source) at
com.jaspersoft.jasperserver.rest.services.RESTResource.doPut(RESTResource.java:220) at
com.jaspersoft.jasperserver.rest.RESTAbstractService.execute(RESTAbstractService.java:113) ...

In the attached files there are two clients create in JAVA and C#, both cause the same error. What is interesting, if I send the seme resource descriptor by PUT method, but without multipart, every thing goes fine. Unfortunately in the future I will need the request with multipart, because I want to create resource with file from remote system (I can not use SOAP based WS).

Best Regards

[christophe_1]

Hi i had the same issue. The documentation is not true.

To create a new folder, dont use multipart/form-data but just text/plain. see my example below :

in this example body=data

put in this url = http://localhost:8080/jasperserver/rest/resource

headers =

     Cookie : JSESSIONID=9D19A14DB810A623C9475673A86A5BEE; Path=/jasperserver

     content-type : text/plain

     User-Agent : JasperServer-Python

data =

 your XML resource descriptor

 

 

 

Reponse =

     status : 201

     content-length : 0

     expires : Thu, 01 Jan 1970 01:00:00 CET

     server : Apache-Coyote/1.1

     pragma : No-cache

     cache-control : no-cache

     date : Thu, 13 Dec 2012 09:54:01 GMT

     p3p : CP="ALL"

     content-type : text/xml;charset=UTF-8

[mateusz.moska]

 

Hi Chrispohe_1,

Thanks for the response, but I think you don't understand me. Creating a folder was just a simple example, in the future I want to create more complicated resources like, jrxml files, images ...

The documentation said: "If the resource has one or more file resources, they must be provided using a multipart request." So I think this is only one posibility to put resourceDescriptor and file content together by REST webservice.

 

 

[christophe_1]

Sorry for the empty post above.

Hey,

To put or post a resource including resourcedescriptor and file resource, see my example below

put in this url = http://localhost:8080/jasperserver/rest/resource

headers =

     Cookie : JSESSIONID=96F74EFB25567E3F0CD872652EAF3FBB; Path=/jasperserver

     Content-Length : 1649

     Content-Type : multipart/form-data; boundary=643e0c37cf39452eb005e5be13a4da72

     User-Agent : JasperServer-Python

data =

--643e0c37cf39452eb005e5be13a4da72

Content-Disposition: form-data; name="ResourceDescriptor"

Content-Type: text/plain; charset=utf-8

 

your xml resourcedescriptor with property has data set to true

 

--643e0c37cf39452eb005e5be13a4da72

Content-Disposition: form-data; name="/images/imagetest.png"; filename="/var/lib/tomcat6/webapps/jasperserver/wcf/tree/leaf.png"

Content-Type: image/png

 

--643e0c37cf39452eb005e5be13a4da72--

 

Reponse =

     status : 201

     content-length : 0

     expires : Thu, 01 Jan 1970 01:00:00 CET

     server : Apache-Coyote/1.1

     pragma : No-cache

     cache-control : no-cache

     date : Thu, 13 Dec 2012 12:30:09 GMT

     p3p : CP="ALL"

     content-type : text/xml;charset=UTF-8

 

the name of the second argument must be the same as the uriString

 

[mateusz.moska]

Hi,

The answer was very helpful, thanks a lot!
The example in documentation is incorrect. You can't create resource using multipart content-type without sending a file content.

In my example you can create folder using multipart content type, but you have to add dummy file as one part in the request.

[christophe_1]

I know, I told you the documentation is false and incomplete

FYI, i found a good code example lost somewhere in this website : http://community.jaspersoft.com/wiki/getting-started-rest-web-service-api

you'll find java code example at the bottom of page. Just read the code in the following path :

codeexamplesrctestjavacomjaspersoftjasperserverrestsampleBasicResourceCRUDTest.java

I hope i help a lot of people, because i never found any informations about creating resource with RESTapi

[anjalisaxena]

When I am trying to follow the above example I am getting :HTTP/1.1 403 Forbidden.Can you please share the format of your resource descriptor. Appreciate your help in advance.

[christophe_1]

Hi, if you get a 403 that it's not a Resource Descriptor format issue. You try to use method not allowed. Please explain you're trying to do.

Christophe

[anjalisaxena]

Hi Chris, Appreciate your response.

I have copied a jrxml from the server by GET method and now I am trying to put that back to the same location meaning I need to use addormodify, in REST API as per documentation I have to use POST method with resource service. I created a multi part request as per the above example. I am pasting the pseudo-code below

MultipartEntity multipartEntity = null;

HttpPost generateReport = new HttpPost("http://localhost:8070/jasperserver/rest/resource/pathtoresource");

multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);

multipartEntity.addPart("ResourceDescriptor", new StringBody(requestXml));

multipartEntity.addPart("http://localhost:8070/jasperserver/rest/resource/pathtoresource", new FileBody(new File("location of jrxml")));

generateReport.setEntity(multipartEntity);

HttpResponse httpResponse = httpClient.execute(generateReport, localContext);

 

Before this code I am calling the login function an I am getting status 200 but then 403.

Anjali

[christophe_1]

You say : "I have copied a jrxml from the server by GET method and now I am trying to put that back to the same location meaning I need to use addormodify, in REST API as per documentation I have to use POST method with resource service"

 

Exact !

 

You say : "http://localhost:8070/jasperserver/rest/resource/pathtoresource"

the documentation is not exact.

You have to use pathtoresource for PUT method BUT pathtoresource/resourcetomodify for POST method.

 

You can't use POST in a folder. You need to know name the of resource you want to modify.

 

Do you understand i mean ?

[anjalisaxena]

Thanks for your reply. Exactly that's what I am trying to do. I am doing a POST and then in the above statement "http://localhost:8070/jasperserver/rest/resource/pathtoresource" here actually path to resource is the path to the reportunit, which is what I need to modify.

Similarly in the same API I have a need to copy a reportunit from one folder to another within the jasperserver. For that I am using PUT which is giving me 400 error.

[anjalisaxena]

One more thing I will like to share, the way I am trying to capture cokkies

 

DefaultHttpClient httpClient = new DefaultHttpClient();

HttpContext localContext = new BasicHttpContext();

 

 

try{

HttpClient client = new HttpClient();

 

// Setting Login URL in a POST method

String loginURL = "http://localhost:8070/jasperserver/rest/login";

HttpPost loginCall = new HttpPost("http://localhost:8070/jasperserver/rest/login");

PostMethod postMethod = new PostMethod(loginURL);

httpClient.getCredentialsProvider().setCredentials(

new AuthScope("http://localhost", 8070),

new UsernamePasswordCredentials("username", "password"));

 

cookieStore = new BasicCookieStore();

// Set authentication parameters

postMethod.addParameter("j_username", "username");

postMethod.addParameter("j_password", "password");

int statusCode = client.executeMethod(postMethod);

if (statusCode != HttpStatus.SC_OK) {

System.out.println("Login failed: " + postMethod.getStatusLine());

return;

}else

System.out.println("The value of login status is:"+postMethod.getStatusLine());

 

 

// Create local HTTP context

// Bind custom cookie store to the local context

localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);

 

loginCall.setHeader("content-type", "application/x-www-form-urlencoded");

 

[christophe_1]

error 400 means BadRequest : check format of your resourceDescriptor.

Why dont' use the web interface to copy ?

 

[anjalisaxena]

This is the need of my application. We need to copy the jrxml and then change it on local and deploy it back and then generate the reports.

Main thing is the issue of 403 on modifying.Any pointers for that.

[christophe_1]

did you add the resource name in your path as '/pathtoresource/resourcename' ?

403 means the method is not allowed.

 

 

[anjalisaxena]

Yes I did added.

[christophe_1]

FYI , it exists several REST clients addons in web browser (Chrome and firefox).

Go to chrome webstore and look for 'REST' term.

My favorite is Postman REST. You'll be able to try the API faster than modify your code.

I spent a lot of time on this API. This tools is really efficient to test the API, and writing the code.

[anjalisaxena]

Already using cREST client plugin in CHROME. Same error. Thanks for your help though.