REST服务返回HTTP 204(Tomcat/Linux)

Question

我在linux发行版(Raspbian,Tomcat 7)上运行代码时遇到问题.该问题未出现在Windows/Tomcat 7/Eclipse下的测试环境中:

我的webservice只返回HTTP 204而没有别的.该日志不包含请求期间发生错误的任何迹象.它应该像在测试环境中那样用json响应.

web.xml中

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<display-name>***</display-name>
<welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<listener>
    <listener-class>**.useravailability.UserAvailabilityServletContextListener</listener-class>
</listener>
<servlet-mapping>
    <servlet-name>REST Service</servlet-name>
    <url-pattern>/REST/*</url-pattern>
</servlet-mapping>
<servlet>
    <servlet-name>REST Service</servlet-name>
    <servlet-class>com.sun.jersey.server.impl.container.servlet.ServletAdaptor</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
        <param-value>true</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

</web-app>

usersService.xml

package **.webservice;

import java.util.ArrayList;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import com.google.gson.Gson;

import **.model.AccessManager;

@Path("/usersService")
public class UsersService {
@GET
@Produces("application/json")
public String getUser() {       
    String user = null;
    ArrayList<String> userList = new ArrayList<String>();
    try {       
        userList = new AccessManager().getUser();
        Gson gson = new Gson();
        user = gson.toJson(userList);           
    } catch (Exception e) {
        e.printStackTrace();
    }
    return user;
}


}

我错过了什么？为什么它在测试环境中有效？谢谢!

Answer 1

如果返回null,Jersey返回HTTP 204.

我认为这是最好的行为.

在您的情况下,您可能会引发异常,因此user为null.

如果要返回错误代码(可能更符合逻辑),可以替换

catch (Exception e) {
        e.printStackTrace();
    }

通过

catch (Exception e) {
        throw new WebApplicationException(404);
    }

如果你想说明找不到用户,或者你想要返回服务器端错误代码的5XX HTTp.