jersey

I developed a Maven Web Project to expose some jersey web services through Payara 4, the project builds fine, but my web methods are not exposed. Here is the main class: package br.com.ainstec.cfrjob.api.resources; import br.com.concil.cfrimporter.controller.CfrImport; import br.com.concil.cfrimporter.model.PojoTest; import javax.enterprise.context.RequestScoped; import javax.ws.rs.*; import javax.ws.rs.core.MediaType; import org.slf4j.Logger; import org.slf4j.LoggerFactory; import org.springframework.context.support.ClassPathXmlApplicationContext; /** * Recurso REST para tratar chamadas * */

I am officially perplexed by this blocker. I had upgraded Jersey from 2.4.1 to 2.5.1 because Jersey/Oracle fixed a package scanning bug that was affecting getting Swagger working. 2.4.1 was working relatively "okay" after lots of hacks to integrate with our Spring-based server (spring-jersey3 was not working at all .) Now since upgrading to 2.5.1 (and now, 2.6-SNAPSHOT), I receive the error below each time I make a request to anything matching the API URL pattern. What I have done: Redeployed on a new, clean server (clean maven repository, clean tomcat, etc); Checked that the jar is the

I'm using Jersey 1.18 and I have POJOMappingFeature enabled. I would like all incoming & outgoing JSON values to be converted from null to an empty string. How to configure the mapping? I've searched web but was unable to find what I'm looking for. from my web.xml: <servlet> <servlet-name>Jersey REST Service</servlet-name> <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class> <init-param> <param-name>com.sun.jersey.config.property.packages</param-name> <param-value>com.abc.restControllers</param-value> </init-param> <init-param> <param-name>com.sun.jersey.api

直到 Jersey 2.6 版本，Jersey 将会兼容 Java SE 6。这个情况将会在在 Jersey 2.7 的版本有所改变。 直到 Jersey 版本 2.25x， 所有的 Jersey 的组件将会兼容 Java SE 7 为目标。这个意思是你将会需要最少使用 Java SE 7 来支持和运行使用最新的。只有 core-common 和 core-client 模块仍然兼容 Java SE 6 的版本并且可以在这个版本上运行。 从 Jersey 2.26 开始，所有的模块将会使用 Java SE 8 进行构建，这里并不能支持运行老的 Java SE 分发包。 https://www.cwiki.us/display/JERSEYZH/Java+SE+Compatibility 来源： oschina 链接： https://my.oschina.net/u/2344080/blog/2871538

I am trying my first REST application. Below is my configuration: Java 7 JBoss AS 7.1 Jersey 2.22.2 Below are list of jars downloaded by Maven: <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>org.prasad</groupId> <artifactId>messenger</artifactId> <packaging>war</packaging> <version>0.0.1-SNAPSHOT</version> <name>messenger</name> <build> <finalName>messenger</finalName> <plugins> <plugin> <groupId

I'm new here even though I've found many answers to my problems in here before. Now I'm looking for help with this: I have this little example resource on my little REST API: @Path("/greeting") @PermitAll public class HelloResource { @GET @Produces(MediaType.TEXT_PLAIN) @Path("all") public String sayHelloToAll() { return "Hello, everybody!"; } @GET @Produces(MediaType.TEXT_PLAIN) @RolesAllowed("admin") @Path("admin") public String sayHelloToAdmin() { return "Hello, admin!"; } } In order to filter roles, I have this implementation of SecurityContext: public class Authorizer implements