Java Rest Service XML 请求语法不正确

Question

我是网络服务的新手。我正在构建一个 android 应用程序，我正在尝试向 Web 服务发送一个发布请求，但我不确定正确的格式是什么。

我能否以 XML 类型发送它，我是否必须修改 POST 方法才能从我的应用程序访问它？

这是 REST 中的 POST 方法：

@POST
@Override
@Consumes({MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON,MediaType.TEXT_XML})
public void create(Users entity) {
    super.create(entity);
}

这是与我们的表对应的类。我正在使用 Oracle 11g 数据库、Glassfish 服务器 4.1.1 和 Netbeans。我正在尝试从我的 android 应用程序中调用 PUT。有人可以建议一种方法吗？

import java.io.Serializable;
import javax.persistence.Basic; 
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.OneToOne;
import javax.persistence.Table;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;
import javax.xml.bind.annotation.XmlRootElement;

@Entity
@Table(name = "USERS")
@XmlRootElement
@NamedQueries({
@NamedQuery(name = "Users.findAll", query = "SELECT u FROM Users u"),
@NamedQuery(name = "Users.findByUserId", query = "SELECT u FROM Users u WHERE u.userId = :userId"),
@NamedQuery(name = "Users.findByUsername", query = "SELECT u FROM Users u WHERE u.username = :username"),
@NamedQuery(name = "Users.findByPassword", query = "SELECT u FROM Users u WHERE u.password = :password"),
@NamedQuery(name = "Users.findByEmail", query = "SELECT u FROM Users u WHERE u.email = :email")})


public class Users implements Serializable {

private static final long serialVersionUID = 1L;
@Id
@Basic(optional = false)
@NotNull
@Column(name = "USER_ID")
private Short userId;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 30)
@Column(name = "USERNAME")
private String username;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 20)
@Column(name = "PASSWORD")
private String password;

@Basic(optional = false)
@NotNull
@Size(min = 1, max = 40)
@Column(name = "EMAIL")
private String email;
@OneToOne(cascade = CascadeType.ALL, mappedBy = "users")
private Profiles profiles;

public Users() {
}

public Users(Short userId) {
    this.userId = userId;
}

public Users(Short userId, String username, String password, String email) {
    this.userId = userId;
    this.username = username;
    this.password = password;
    this.email = email;
}

public Short getUserId() {
    return userId;
}

public void setUserId(Short userId) {
    this.userId = userId;
}

public String getUsername() {
    return username;
}

public void setUsername(String username) {
    this.username = username;
}

public String getPassword() {
    return password;
}

public void setPassword(String password) {
    this.password = password;
}

public String getEmail() {
    return email;
}

public void setEmail(String email) {
    this.email = email;
}

public Profiles getProfiles() {
    return profiles;
}

public void setProfiles(Profiles profiles) {
    this.profiles = profiles;
}

@Override
public int hashCode() {
    int hash = 0;
    hash += (userId != null ? userId.hashCode() : 0);
    return hash;
}

@Override
public boolean equals(Object object) {
    // TODO: Warning - this method won't work in the case the id fields are not set
    if (!(object instanceof Users)) {
        return false;
    }
    Users other = (Users) object;
    if ((this.userId == null && other.userId != null) || (this.userId != null && !this.userId.equals(other.userId))) {
        return false;
    }
    return true;
}

@Override
public String toString() {
    return "medicalplanner.Users[ userId=" + userId + " ]";
}

}

Answer 1

您可以为此使用 REST 或 SOAP 协议。 JSON 格式比这更合适。如果您需要更高的安全性，请使用 SOAP。但是 SOAP 比 JSON 格式慢。你可以在这里看到不同之处：
SOAP vs REST (differences)
还有一些用于创建 JSON 或 XML 格式的类的库。我建议杰克逊图书馆。 https://github.com/FasterXML/jackson 。您也可以查看 GSON 库。 https://github.com/google/gson

Answer 2

您绝对可以按照您列出的方式进行操作 - 我有 JAX-RS 应用程序，它们可以采用 JSON 或 XML。

你能告诉我们更多关于错误的信息吗？老实说，听起来您的 XML 中有一个错误，这使解析器感到困惑。您可以粘贴您发送的 XML。