Question

@Path("file.upload")
public class UploadFileService {
@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
        @FormDataParam("file") InputStream uploadedInputStream,
        @FormDataParam("file") FormDataContentDisposition fileDetail) {

    System.out.println("-----------------UploadFileService------------------1.");
    // Should we use a disk or DB? Decided to use DISK
    // Path should be read from properties-files
    String uploadedFileLocation = "//uploaded/" + fileDetail.getFileName();

    // save it
    writeToFile(uploadedInputStream, uploadedFileLocation);

    String output = "File uploaded to : " + uploadedFileLocation;
    // All went OK
    return Response.status(200).entity(output).build();

}

 WARNING: No injection source found for a parameter of type public javax.ws.rs.core.Response com.insame.service.UploadFileService.uploadFile(java.io.InputStream,com.sun.jersey.core.header.FormDataContentDisposition) at index 0.

SEVERE:   WebModule[/insame]StandardWrapper.Throwable
org.glassfish.jersey.server.model.ModelValidationException: Validation of the application resource model has failed during application initialization.
[[FATAL] No injection source found for a parameter of type public javax.ws.rs.core.Response com.insame.service.UploadFileService.uploadFile(java.io.InputStream,com.sun.jersey.core.header.FormDataContentDisposition) at index 0.; source='ResourceMethod{httpMethod=POST, consumedTypes=[multipart/form-data], producedTypes=[], suspended=false, suspendTimeout=0, suspendTimeoutUnit=MILLISECONDS, invocable=Invocable{handler=MethodHandler{handlerClass=class com.insame.service.UploadFileService, handlerConstructors=[org.glassfish.jersey.server.model.HandlerConstructor@47bee27a]}, handlingMethod=public javax.ws.rs.core.Response com.insame.service.UploadFileService.uploadFile(java.io.InputStream,com.sun.jersey.core.header.FormDataContentDisposition), parameters=[Parameter [type=class java.io.InputStream, source=file, defaultValue=null], Parameter [type=class com.sun.jersey.core.header.FormDataContentDisposition, source=file, defaultValue=null]], responseType=class javax.ws.rs.core.Response}, nameBindings=[]}']
    at org.glassfish.jersey.server.ApplicationHandler.initialize(ApplicationHandler.java:410)
    at org.glassfish.jersey.server.ApplicationHandler.access$500(ApplicationHandler.java:157)
    at org.glassfish.jersey.server.ApplicationHandler$3.run(ApplicationHandler.java:280)
    at org.glassfish.jersey.internal.Errors$2.call(Errors.java:289)
    at org.glassfish.jersey.internal.Errors$2.call(Errors.java:286)
    at org.glassfish.jersey.internal.Errors.process(Errors.java:315)
    at org.glassfish.jersey.internal.Errors.process(Errors.java:297)
    at org.glassfish.jersey.internal.Errors.processWithException(Errors.java:286)
    at org.glassfish.jersey.server.ApplicationHandler.<init>(ApplicationHandler.java:277)
    at org.glassfish.jersey.servlet.WebComponent.<init>(WebComponent.java:262)
    at org.glassfish.jersey.servlet.ServletContainer.init(ServletContainer.java:167)

我在UploadFileService：

下实现了这样的测试服务

@GET
@Path("count")
@Produces("text/plain")
public String countREST() {
    return "1 one 1";
}

我记录了这个例外：

FINE:   [Web-Security] hasResource perm: ("javax.security.jacc.WebResourcePermission" 

"/webresources/file.upload/count" "GET")
SEVERE:   WebModule[/insame]StandardWrapper.Throwable
java.lang.IllegalStateException: The resource configuration is not modifiable in this context.
    at org.glassfish.jersey.server.ResourceConfig$ImmutableState.register(ResourceConfig.java:257)
WARNING:   StandardWrapperValve[com.insame.service.ApplicationConfig]: Allocate exception for servlet com.insame.service.ApplicationConfig
java.lang.IllegalStateException: The resource configuration is not modifiable in this context.
    at org.glassfish.jersey.server.ResourceConfig$ImmutableState.register(ResourceConfig.java:257)
    at org.glassfish.jersey.server.ResourceConfig$ImmutableState.register(ResourceConfig.java:205)
    at org.glassfish.jersey.server.ResourceConfig.register(ResourceConfig.java:435)
    at org.glassfish.jersey.servlet.WebComponent.<init>(WebComponent.java:261)
    at org.glassfish.jersey.servlet.ServletContainer.init(ServletContainer.java:167)
    at org.glassfish.jersey.servlet.ServletContainer.init(ServletContainer.java:349)

环境

Netbeans7.3.1
Glassfish 4.0
使用Glassfish 4.0的泽西岛2

Answer 1

要在Jersey应用程序中使用multipart，您需要在应用程序中注册MultiPartFeature，即：

public class ApplicationConfig extends Application {

    public Set<Class<?>> getClasses() {
        final Set<Class<?>> resources = new HashSet<Class<?>>();

        // Add your resources.
        resources.add(UploadFileService.class);

        // Add additional features such as support for Multipart.
        resources.add(MultiPartFeature.class);

        return resources;
    }
}

有关详细信息，请参阅“泽西用户指南”中的Multipart部分。

对于您正面临的第二个问题，尝试重新启动GlassFish服务器，我不确定NetBeans在更改后如何重新加载Jersey应用程序（如果这没有帮助，请发布您的ApplicationConfig）。

Answer 2

我遇到了同样的问题，并希望避免创建自定义应用程序类。它没有很好的文档记录，但是如果你想添加Multipart功能，你所要做的就是将它添加到你的web.xml jersey servlet config：

<init-param>
    <param-name>jersey.config.server.provider.classnames</param-name>
    <param-value>org.glassfish.jersey.filter.LoggingFilter;org.glassfish.jersey.media.multipart.MultiPartFeature</param-value>
</init-param>

我还添加了一个loggingfilter。

Answer 3

如果您正在使用jetty服务器和jersey servlet，那么您可以通过在启动了jetty服务器的主类中添加以下代码来解决此问题，

ServletHolder jerseyServlet = context.addServlet（ org.glassfish.jersey.servlet.ServletContainer.class，“/ *”）; jerseyServlet.setInitOrder（0）;
      // Tells the Jersey Servlet which REST service/classes to load.
      jerseyServlet
              .setInitParameter(
                      "jersey.config.server.provider.classnames",
                      <Your entry point class's canonical name>
                              + ";org.glassfish.jersey.media.multipart.MultiPartFeature");

Answer 4

稍作澄清

使用

import org.glassfish.jersey.media.multipart.MultiPartFeature
import org.glassfish.jersey.media.multipart.FormDataContentDisposition;
import org.glassfish.jersey.media.multipart.FormDataParam;

不

com.sun.jersey.*

仅在使用org.glassfish.jersey.media.multipart.*

时才对我有效

在ApplicationConfig中，只需将MultiPartFeature注册为

import org.glassfish.jersey.media.multipart.MultiPartFeature;

@javax.ws.rs.ApplicationPath("webresources")
public class ApplicationConfig extends Application {

    @Override
    public Set<Class<?>> getClasses() {
        Set<Class<?>> resources = new java.util.HashSet<>();
        resources.add(UploadFileService.class);
        resources.add(MultiPartFeature.class);
        return resources;
    }
}

Answer 5

我使用Jersey 1.9.1。 org.glassfish ......适用于Jersey 2.对于Jersey 1，你最好使用com.sun ... classes。

Answer 6

如果您希望使用注释，可以使用与FormDataMultiPart等效的@FormDataParam（＆＃34; file＆＃34;）。

使用以下给出的示例代码提取：

public Response uploadFile( **@FormDataParam("file")** InputStream fileInputStream,
             @FormDataParam("file") FormDataContentDisposition contentDispositionHeader) {

使用Jersey通过RESTful服务上传文件和资源配置是不可修改的？

6 个答案: