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[英]deserialise the list of integer from json to java using Jackson Object Mapper
[英]Jackson object mapper not creating proper Json string from java object
我的密钥中没有 A、B、C 我有 user_id、密码等密钥。此外,如果我打印用户 class 的 toString,它会显示正确的内容。
注意:用户 object 来自 sqlite 数据库,我仔细检查了它是否工作正常。
请同时检查附加用户 class。 我用这个 class 来保存 sqlite 值。
So I am getting user data from sqlite in the form of user object then converting user object into jsonstring using Jackson object mapper.
如果您指出问题所在,那将是非常有帮助的。
目前我低于 jsnstring,这是错误的。
{"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 object 转换时,我在 jsonstring 中得到 abcd “key”而不是实际密钥。
我的实际密钥就像 user_id、email 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;
}
public class User {
public int userid;
public String email;
public String updatedemail;
public String firstname;
public String lastname;
// Empty constructor
public User() {
}
// constructor
public User(int userid, String email, String updatedemail, String firstname, String lastname){
this.userid = userid;
this.email = email;
this.updatedemail = updatedemail;
this.firstname = firstname;
this.lastname = lastname;
}
public int getID() {
return this.userid;
}
public void setID(int userid) {
this.userid = userid;
}
public String getemail() {
return this.email;
}
public String getfirstname() {
return this.firstname;
}
public String getlastname() {
return this.lastname;
}
public void setFirstName(String firstname) {
this.firstname = firstname;
}
public void setLastName(String lastname) {
this.lastname = lastname;
}
public void setEmail(String email) {
this.email = email;
}
public String getNewemail() {
return this.updatedemail;
}
public void setNewemail(String newemail) {
this.updatedemail = newemail;
}
@Override
public String toString() {
return "User{" +
"userid=" + userid +
", email='" + email + '\'' +
", updatedemail='" + updatedemail + '\'' +
", firstname='" + firstname + '\'' +
", lastname='" + lastname + '\'' +
'}';
}
}
这是 toString() 的 output
User{userid=91316, email='vasimflutter2@gmail.com', updatedemail='', firstname='Vasim', lastname='Flutter2'}
您的User
POJO 应正确注释:
@JsonProperty("key of your element in JSON")
进行注释(例如@JsonProperty("userId")
字段private int userId
;@JsonCreator
告诉 Jackson 它应该在构建 object 时使用该构造函数@JsonProperty(name = "the key of your element", required = true/false)
进行注释getElement()
- 您可以使用 IDE 自动创建它们正确注释 POJO 后,您可以:
objectMapper.valueToTree(jsonString, User.class)
:objectMapper.writeValueAsString(user)
创建现有User user
实例的 JSON 表示。注意:注释不是“强制性的”,但强烈推荐。 如果你不注释你的字段和构造函数/getter,Jackson 将不得不猜测。 对库进行猜测从来都不是一件好事,最好是明确的,这样您就可以根据需要命名属性,或者可能在没有副作用的情况下出错。
总结一下:
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;
}
}
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