Jackson对象映射器未从java对象创建正确的Json字符串
我的密钥中没有A、B、C,我有用户id、密码等密钥。 另外,如果我打印用户类的toString,它会显示正确的内容 注意:用户对象来自sqlite数据库,我仔细检查它是否工作正常 请检查附加用户类。我使用这个类来保存sqlite值 因此,我以用户对象的形式从sqlite获取用户数据,然后使用Jackson对象映射器将用户对象转换为jsonstring 如果你能指出问题所在,那将大有帮助 目前,我正在低于jsnstring,这是错误的Jackson对象映射器未从java对象创建正确的Json字符串,java,android,sqlite,jackson,jackson-databind,Java,Android,Sqlite,Jackson,Jackson Databind,我的密钥中没有A、B、C,我有用户id、密码等密钥。 另外,如果我打印用户类的toString,它会显示正确的内容 注意:用户对象来自sqlite数据库,我仔细检查它是否工作正常 请检查附加用户类。我使用这个类来保存sqlite值 因此,我以用户对象的形式从sqlite获取用户数据,然后使用Jackson对象映射器将用户对象转换为jsonstring 如果你能指出问题所在,那将大有帮助 目前,我正在低于jsnstring,这是错误的 {"A":1,"B
{"A":1,"B":1,"C":1,"D":1,"E":1,"F":1,"G":0,"H":"ef","I":"","J":0,"K":30,"L":"","M":1,"N":0,"O":1,"P":0,"Q":"","R":"","S":1,"T":"","U":1,"V":"fkGinMCh02k:APA91bH1I8Hv1EIGdkRtZiqjvsMY-ixk_crcNxSQ3gQ1PuAOoQ0B4qllstCdOw43nZQ90JiqTpcfCyQ-_y6RsxnWcjg0gojqZ8pv4Fia_9mW4-De7nPQF4C5XIF16V5","W":"","X":0,"Y":"1","Z":"1,2,3,4,5,6,7,9,16","a":91316,"b":"email@gmail.com","c":"","d":"im","e":"F","f":"","g":1,"h":1,"i":1,"j":2478,"k":492372,"l":0,"m":10,"n":0,"o":"","p":0,"q":0,"r":0,"s":91316,"t":1,"u":1,"v":1,"w":1,"x":1,"y":0,"z":0}
预期的jsonstring
{"user_id":1,"first_name":"test","last_name":"test"}
当我从java对象转换时,我在jsonstring中得到的是“key”,而不是实际的key
我的实际密钥类似于用户id、电子邮件等
private String getUserObjectString() {
String jsonStr = "";
User user = here I am getting actual object content;
ObjectMapper mapper = new ObjectMapper();
try {
jsonStr = mapper.writeValueAsString(user);
} catch (JsonProcessingException e) {
Log.i("Exception","1");
e.printStackTrace();
}
return jsonStr;
}
这是toString()的输出
您的
用户
POJO应正确注释:
@JsonProperty(“JSON中元素的键”)
注释(例如@JsonProperty(“userId”)
字段私有int userId
@JsonCreator
,告诉Jackson在构建对象时应该使用该构造函数@JsonProperty(name=“元素的键”,required=true/false)注释
getElement()
——您可以使用IDE自动创建它们objectMapper.valueToTree(jsonString,User.class)
objectMapper.writeValueAsString(User)
创建现有用户
实例的JSON表示总而言之:
public class User {
@JsonProperty("userId") // <- note: the literal value here should be exactly what you see in your Json (case included)
public int userid;
@JsonProperty("email")
public String email;
@JsonProperty("updateEmail")
public String updatedemail;
@JsonProperty("firstName")
public String firstname;
@JsonProperty("lastName")
public String lastname;
@JsonCreator
public User(
@JsonProperty(name = "userId", required = true) int userid,
@JsonProperty(name = "email", required = true) String email,
@JsonProperty(name = "updateEmail", required = true) String updatedemail,
@JsonProperty(name = "firstName", required = true) String firstname,
@JsonProperty(name = "lastName", required = true) String lastname
) {
this.userid = userid;
this.email = email;
this.updatedemail = updatedemail;
this.firstname = firstname;
this.lastname = lastname;
}
public int getUserid() {
return userid;
}
public String getEmail() {
return email;
}
public String getUpdatedemail() {
return updatedemail;
}
public String getFirstname() {
return firstname;
}
public String getLastname() {
return lastname;
}
}
公共类用户{
@JsonProperty(“userId”)//不确定这是否是问题的原因,但getter和setter的命名已关闭。要让jackson使用访问器工作,getter和setter必须严格遵守bean命名约定:getAttributeName
(其中,attributeName
是对应属性的名称。get
之后的第一个字母始终大写。此外,属性应设置为private
,而不是public
).@Turing85是的,但那将是在我们创建json到java对象时发生的,但在这里我将java对象转换为jsonstring。两种方式(序列化和反序列化)都使用bean约定。@Turing85好的,让我试试完美名称和私有修饰符,然后再回复您。谢谢
User{userid=91316, email='vasimflutter2@gmail.com', updatedemail='', firstname='Vasim', lastname='Flutter2'}
public class User {
@JsonProperty("userId") // <- note: the literal value here should be exactly what you see in your Json (case included)
public int userid;
@JsonProperty("email")
public String email;
@JsonProperty("updateEmail")
public String updatedemail;
@JsonProperty("firstName")
public String firstname;
@JsonProperty("lastName")
public String lastname;
@JsonCreator
public User(
@JsonProperty(name = "userId", required = true) int userid,
@JsonProperty(name = "email", required = true) String email,
@JsonProperty(name = "updateEmail", required = true) String updatedemail,
@JsonProperty(name = "firstName", required = true) String firstname,
@JsonProperty(name = "lastName", required = true) String lastname
) {
this.userid = userid;
this.email = email;
this.updatedemail = updatedemail;
this.firstname = firstname;
this.lastname = lastname;
}
public int getUserid() {
return userid;
}
public String getEmail() {
return email;
}
public String getUpdatedemail() {
return updatedemail;
}
public String getFirstname() {
return firstname;
}
public String getLastname() {
return lastname;
}
}