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manipulate JSON array in Spring Hibernate


manipulate JSON array in Spring Hibernate

By : user2953899
Date : November 21 2020, 07:31 AM
To fix the issue you can do I'd use transient variables as a solution for this problem. I don't think there is a better solution for this. You may also try hibernate interceptors for this as a solution.
code :
@Table
@Entity(name='Likes')
public class LikesEntity implements Serializable {

    @Transient
    private List<String> userIds;
    @Column
    private String likes;
    @Column
    private String count;

     //getter and setters for the like and count
    ...
    public void addUserId(String id){
        this.userIds.add(userIds);
        this.likes = convertToJsonArray(userIds); // use GSON or Jackson or any other library that can help you convert array to JSON string
        this.count = this.userIds.size();
    } 


}
 @Transactional(lockmode = LockMode.READ){
   LikesEntity le = sess.get(LikesEntity.class,234);
   le.addUserId("userPk");
   session.saveOrUpdate(le);
 }


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How to manipulate this JSON array in JavaScript

How to manipulate this JSON array in JavaScript


By : Bartimoot
Date : March 29 2020, 07:55 AM
To fix the issue you can do That's not JSON, that's a Javascript object. JSON is a text format for representing data.
First you would need to find Jake. He is in the data.children array, so look there:
code :
var i = 0;
while (data.children[i].name != 'Jake') i++;
data.children[i].children = [ 
  { name: 'Angela', level: 3 }   
];
Mapping JSON Array to Spring Hibernate Database (H2)

Mapping JSON Array to Spring Hibernate Database (H2)


By : Surbhi Dhingra
Date : March 29 2020, 07:55 AM
I wish did fix the issue. If I understand your requirement, you want to persist this array private String[] foodPairing; in one column. If so, you can proceed like below :
code :
public class User {
   ....
   @Transient 
   private String[] foodPairing;

   @Column(name="food_pairing")
   private String food_Pairing;
}
user.setFood_Paring(user.getFoodPairing().toString());
public class User {
   ....
   @Transient 
   private String[] foodPairing;

   @JsonIgnore
   @Column(name="food_pairing1")
   private String foodPairing1;

   @JsonIgnore
   @Column(name="food_pairing2")
   private String foodPairing2;

   @JsonIgnore
   @Column(name="food_pairing3")
   private String foodPairing3;
}
user.setFoodParing1(user.getFoodPairing()[0]); //etc...
Manipulate an array or json reponse and save it as another array in javascript

Manipulate an array or json reponse and save it as another array in javascript


By : user2400199
Date : March 29 2020, 07:55 AM
hop of those help? You can use Reduce
code :
let data = [{"Latitude":"17.4283466666667","Longitude":"78.5797422222222", "VehicleNumber":"TS08UA6743"},{"Latitude":"17.4282822222222","Longitude":"78.5799111111111","VehicleNumber":"TS08UA6743"},{"Latitude":"17.4284233333333","Longitude":"78.5797333333333","VehicleNumber":"TS08UA6745"}]

let op = data.reduce((out,{VehicleNumber:vn,Latitude:lat,Longitude:lon})=>{
  if(out[vn]){ out[vn].push([lat,lon]) }
  else { out[vn] = [[lat,lon]] }
  return out
},{})

console.log(op)
((out[vn] ? out[vn].push([lat,lon]) : out[vn]=[[lat,lon]]), out)
How to store JSON file as a JSON datatype in mysql using Hibernate / Spring boot?

How to store JSON file as a JSON datatype in mysql using Hibernate / Spring boot?


By : user3628121
Date : March 29 2020, 07:55 AM
hope this fix your issue Your code reads json into list of User objects and persists to database. You need to write some custom logic to save it as json. There are multiple ways to do it.
You can do something like
code :
@NonNull
@Column(columnDefinition = "JSON") // Only if Database has JSON type else this line can be removed
private String jsonData;
@PrePersist
public void prePersist() {
    try {
        this.setJsonData(new ObjectMapper().writeValueAsString(this));
    } catch (JsonProcessingException e) {
        e.printStackTrace();
    }
}
public void setJsonData(String jsonData) {
    // Method parameter jsonData is simply ignored
    try {
        this.jsonData = new ObjectMapper().writeValueAsString(this);
    } catch (JsonProcessingException e) {
        log.error(e.getMessage());
    }
}
How to manipulate jQuery AJAX JSON data in a controller in spring MVC

How to manipulate jQuery AJAX JSON data in a controller in spring MVC


By : victorvicpal
Date : March 29 2020, 07:55 AM
this will help From Ajax you can also pass data as data:'name='+ name+'&lastname='+ lastname, And at controller end you can make use of @RequestParam annotation to get this value passed from ajax call. Ajax code looks as follows:
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