将原始 JSON 保存为数据库中的字符串

Question

如何使用 POST 请求将原始 Json 保存为 MsSql 数据库中的字符串 - 使用 Jackson ObjectMapper 将字符串转换为 Json 但无法将原始 json 更改为字符串？

{
     "id": 1,
     "someName":"someName",
     "json": {
         "title": "example glossary",
         "GlossDiv": {
             "title": "S",
             "GlossTerm": "Standard Generalized Markup Language"
         }
},
     "anotherjson":{
          "name":"someone",
          "age": 121
},
     "somedate":"03-11-2019.00:00:00" 
}

如何将此保存的 json 保存为数据库中的整数、varchar、字符串、字符串、日期列？ 1,someName, "{"title": "示例词汇表","GlossDiv": {"title": "S","GlossTerm": "标准通用标记语言"}","{"name":"someone" "年龄": 121}", 03-11-2019.00:00:00.

** 更新 ** 为简单起见，这里是简单的 json

{
"id":1,
"jsonObjectHolder":{
        "name": "Name",
        "age" : 404
        }}

控制器：

@PostMapping("/postJson")
public void postJson(@RequestBody TryJson tryJson) {
    tryJsonService.postJson(tryJson);
}

服务：

public void postJson(TryJson tryJson) {
    tryJsonRepository.save(tryJson);
}

回购：

public interface TryJsonRepository extends CrudRepository<TryJson, Integer> {

}

型号：

@Entity

@Table(name = "TryJson") 公共类 TryJson {

@Id
@Column(name = "id")
private Integer id;

@JsonIgnore
@Column(name = "json_column")
private String jsonColumn;

@Transient
private JsonNode jsonObjectHolder;

public TryJson() {
}

public TryJson(Integer id, String jsonColumn) {
    this.id = id;
    this.jsonColumn = jsonColumn;
}

public Integer getId() {
    return id;
}

public void setId(Integer id) {
    this.id = id;
}

public JsonNode getJsonObjectHolder() {
    return jsonObjectHolder;
}

public void setJsonObjectHolder(JsonNode jsonObjectHolder) {
    this.jsonObjectHolder = jsonObjectHolder;
}

public String getJsonColumn() {
    return this.jsonObjectHolder.toString();
}

public void setJsonColumn(String jsonColumn) throws IOException {
    ObjectMapper mapper = new ObjectMapper();
    this.jsonObjectHolder = mapper.readTree(jsonColumn);
}

@Override
public String toString() {
    return String.format("TryJson [Id=%s, JsonColumn=%s, jsonObjectHolder=%s]", id, jsonColumn, jsonObjectHolder);
}

}

http://localhost:8080/api/postJson

ID JSON_COLUMN
1个空

不确定我在这里缺少什么。我确实在调试期间填充了 jsonObjectHolder，但我仍然得到 NULL TryJson [Id=1, JsonColumn=null, jsonObjectHolder={"name":"Name","age":404}]

更新 2

我得到空指针异常。

 Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.orm.jpa.JpaSystemException: Exception occurred inside getter of com.example.tryjson.tryjson.model.TryJson.jsonColumn; nested exception is org.hibernate.PropertyAccessException: Exception occurred inside getter of com.example.tryjson.tryjson.model.TryJson.jsonColumn] with root cause 
java.lang.NullPointerException: null
at com.example.tryjson.tryjson.model.TryJson.getJsonColumn(TryJson.java:52) ~[classes/:na]
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) ~[na:1.8.0_192]

这是我的新模型

@Entity
@Table(name = "TryJson")
public class TryJson {
private Integer id;

@Transient
private JsonNode jsonObjectHolder;

public TryJson() {
}

public TryJson(Integer id, JsonNode jsonObjectHolder) {
    this.id = id;
    this.jsonObjectHolder = jsonObjectHolder;
}

@Id
@Column(name = "id")
public Integer getId() {
    return id;
}

public void setId(Integer id) {
    this.id = id;
}

@Transient
public JsonNode getJsonObjectHolder() {
    return jsonObjectHolder;
}

public void setJsonObjectHolder(JsonNode jsonObjectHolder) {
    this.jsonObjectHolder = jsonObjectHolder;
}

@Column(name = "json_column")
public String getJsonColumn() {
    return this.jsonObjectHolder.toString();
}

public void setJsonColumn(String jsonColumn) throws IOException {
    ObjectMapper mapper = new ObjectMapper();
    this.jsonObjectHolder = mapper.readTree(jsonColumn);
}
}

Answer 1

ObjectMapper mapper = new ObjectMapper();
TypeReference<List<User>> typeReference = new TypeReference<List<Your_Entity>>() {};
InputStream inputStream = TypeReference.class.getResourceAsStream("/bootstrap.json");
try {
    List<Your_Entity> users = mapper.readValue(inputStream, typeReference);
    log.info("Saving users...");
    userService.saveAllUsers(users);
    log.info(users.size() + " Users Saved...");
} catch (IOException e) {
    log.error("Unable to save users: " + e.getMessage());
}

Answer 2

您可以定义一个JsonNode json 属性来保存您想要作为文本持久化的部分，然后将其标记为@Transient，这样JPA 就不会尝试将其存储在数据库中。但是，jackson 应该能够将它来回转换为 Json。

然后您可以为 JPA 编写 getter/setter 代码，这样您就可以从 JsonNode 来回转换为 String。您定义了一个将 JsonNode json 转换为 String 的 getter getJsonString。那个可以映射到一个表列，比如'json_string'，然后你定义一个setter，你从JPA接收字符串并将其解析为JsonNode，这对jackson可用。

不要忘记将@JsonIgnore 添加到getJsonString，这样Jackson 就不会尝试将json 转换为jsonString。

@Entity
@Table(name = "request")
public class Request {

  private Long id;

  private String someName;

  @Transient
  private JsonNode json;

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  public Long getId() {
    return id;
  }

  @Column(name ="someName")
  public String getSomeName() {
    return name;
  }

  public void setSomeName(String name) {
    this.name = name;
  }

  public void setId(Long id) {
    this.id = id;
  }

  // Getter and setter for name

  @Transient  // This is for Jackson
  public JsonNode getJson() {
    return json;
  }

  public void setJson(JsonNode json) {
    this.json = json;
  }


  @Column(name ="jsonString")
  public String getJsonString() { // This is for JPA
    return this.json.toString();
  }

  public void setJsonString(String jsonString) {  // This is for JPA
    // parse from String to JsonNode object
    ObjectMapper mapper = new ObjectMapper();
    try {
      this.json = mapper.readTree(jsonString);
    } catch (Exception e) {
      e.printStackTrace();
    }
  }
}

更新：

如果你用@Column 标记了jsonColumn，spring 将使用反射来提取默认初始化null 的数据，getJsonColumn 永远不会执行翻译：

@JsonIgnore
@Column(name = "json_column")
private String jsonColumn;

你不需要jsonColumn，只要确保你用@Column标记你的setter，所以spring使用gettets/setters来持久化到数据库，持久化时，jpa会执行getJsonColumn，读取时，jpa会执行@ 987654328@ 和 jsonNode 将被来回转换为字符串：

@Entity
@Table(name = "TryJson") public class TryJson {

  private Integer id;

  @Transient
  private JsonNode jsonObjectHolder;

  public TryJson() {
  }

  public TryJson(Integer id, String jsonColumn) {
      this.id = id;
      this.jsonObjectHolder = // use mapper to create the jsonObject;
  }

  @Id
  @Column(name = "id")
  public Integer getId() {
      return id;
  }

  public void setId(Integer id) {
      this.id = id;
  }

  public JsonNode getJsonObjectHolder() {
      return jsonObjectHolder;
  }

  public void setJsonObjectHolder(JsonNode jsonObjectHolder) {
      this.jsonObjectHolder = jsonObjectHolder;
  }

  @Column(name = "json_column")
  public String getJsonColumn() {
      return this.jsonObjectHolder.toString();
  }

  public void setJsonColumn(String jsonColumn) throws IOException {
      ObjectMapper mapper = new ObjectMapper();
      this.jsonObjectHolder = mapper.readTree(jsonColumn);
  }

  @Override
  public String toString() {
      return String.format("TryJson [Id=%s, JsonColumn=%s, jsonObjectHolder=%s]", id, jsonColumn, jsonObjectHolder);
  }
}