我正在尝试通过在 Tomcat 7 上运行 Apache Jersey REST 的 Windows Server 2008 R2 上的 cURL 命令行执行以下命令。
curl -X POST -d "<userList xmlns="urn:user"><user role="ROLE_OPERATOR" loginName="test_login1"></user></userList>" -H "Content-Type: application/xml" --basic --user username:password http://localhost:8080/meolutws/UserList/
当我发出命令时,我收到带有消息的 HTTP 400:“客户端发送的请求在语法上不正确”。
Web 服务上的方法签名如下所示:
@POST
@CONSUMES({"application/xml"})
public Response createUsers(UserList users){
}
UserList 类定义如下:
  @XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "UserList", namespace = "urn:user", propOrder = {
    "users"
})
@XmlRootElement(name = "userList", namespace = "urn:user")
public class UserList
    implements Serializable
{
    @XmlElement(name = "user")
    protected List<User> users;
    public List<User> getUsers() {
        if (users == null) {
            users = new ArrayList<User>();
        }
        return this.users;
    }
}
用户定义如下:
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "User", namespace = "urn:user")
@XmlRootElement(name = "user", namespace = "urn:user")
public class User
    implements Serializable
{
    @XmlAttribute(name = "loginName", required = true)
    protected String loginName;
    @XmlAttribute(name = "role", required = true)
    protected String role;
    public String getLoginName() {
        return loginName;
    }
    public void setLoginName(String value) {
        this.loginName = value;
    }       
    public String getRole() {
        return role;
    }
    public void setRole(String value) {
        this.role = value;
    }
}
我错过了什么?