Java 如何在 JdbcTemplate 中使用 PostgreSQL hstore/json

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时间:2020-08-13 08:24:12  来源:igfitidea点击:

How to use PostgreSQL hstore/json with JdbcTemplate

javaspringpostgresqljdbctemplatehstore

提问by Aymer

Is there a way to use PostgreSQL json/hstore with JdbcTemplate? esp query support.

有没有办法将 PostgreSQL json/hstore 与 一起使用JdbcTemplate?esp 查询支持。

for eg:

例如:

hstore:

存储:

INSERT INTO hstore_test (data) VALUES ('"key1"=>"value1", "key2"=>"value2", "key3"=>"value3"')

SELECT data -> 'key4' FROM hstore_test
SELECT item_id, (each(data)).* FROM hstore_test WHERE item_id = 2

for Json

对于 Json

insert into jtest (data) values ('{"k1": 1, "k2": "two"}');
select * from jtest where data ->> 'k2' = 'two';

采纳答案by sojin

Although quite late for an answer (for the insert part), I hope it might be useful someone else:

虽然答案(对于插入部分)已经很晚了,但我希望它可能对其他人有用:

Take the key/value pairs in a HashMap:

获取 HashMap 中的键/值对:

Map<String, String> hstoreMap = new HashMap<>();
hstoreMap.put("key1", "value1");
hstoreMap.put("key2", "value2");

PGobject jsonbObj = new PGobject();
jsonbObj.setType("json");
jsonbObj.setValue("{\"key\" : \"value\"}");

use one of the following way to insert them to PostgreSQL:

使用以下方法之一将它们插入 PostgreSQL:

1)

1)

jdbcTemplate.update(conn -> {
     PreparedStatement ps = conn.prepareStatement( "INSERT INTO table (hstore_col, jsonb_col) VALUES (?, ?)" );
     ps.setObject( 1, hstoreMap );
     ps.setObject( 2, jsonbObj );
});

2)

2)

jdbcTemplate.update("INSERT INTO table (hstore_col, jsonb_col) VALUES(?,?)", 
new Object[]{ hstoreMap, jsonbObj }, new int[]{Types.OTHER, Types.OTHER});

3) Set hstoreMap/jsonbObj in the POJO (hstoreCol of type Map and jsonbObjCol is of type PGObject)

3)在POJO中设置hstoreMap/jsonbObj(hstoreCol为Map类型,jsonbObjCol为PGObject类型)

BeanPropertySqlParameterSource sqlParameterSource = new BeanPropertySqlParameterSource( POJO );
sqlParameterSource.registerSqlType( "hstore_col", Types.OTHER );
sqlParameterSource.registerSqlType( "jsonb_col", Types.OTHER );
namedJdbcTemplate.update( "INSERT INTO table (hstore_col, jsonb_col) VALUES (:hstoreCol, :jsonbObjCol)", sqlParameterSource );

And to get the value:

并获得价值:

(Map<String, String>) rs.getObject( "hstore_col" ));
((PGobject) rs.getObject("jsonb_col")).getValue();

回答by Vlad Mihalcea

Even easier than JdbcTemplate, you can use the hibernate-typesopen-source project to persist HStore properties.

比 更容易JdbcTemplate,您可以使用hibernate-types开源项目来持久化 HStore 属性。

First, you need the Maven dependency:

首先,您需要 Maven 依赖项:

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-52</artifactId>
    <version>${hibernate-types.version}</version>
</dependency>

Then, assuming you have the following Bookentity:

然后,假设您有以下Book实体:

@Entity(name = "Book")
@Table(name = "book")
@TypeDef(name = "hstore", typeClass = PostgreSQLHStoreType.class)
public static class Book {

    @Id
    @GeneratedValue
    private Long id;

    @NaturalId
    @Column(length = 15)
    private String isbn;

    @Type(type = "hstore")
    @Column(columnDefinition = "hstore")
    private Map<String, String> properties = new HashMap<>();

    //Getters and setters omitted for brevity
}

Notice that we annotated the propertiesentity attribute with the @Typeannotation and we specified the hstoretype that was previously defined via @TypeDefto use the PostgreSQLHStoreTypecustom Hibernate Type.

请注意,我们使用注释对properties实体属性进行了@Type注释,并且我们指定了hstore之前通过@TypeDef使用PostgreSQLHStoreType自定义 Hibernate 类型定义的类型。

Now, when storing the following Bookentity:

现在,当存储以下Book实体时:

Book book = new Book();

book.setIsbn("978-9730228236");
book.getProperties().put("title", "High-Performance Java Persistence");
book.getProperties().put("author", "Vlad Mihalcea");
book.getProperties().put("publisher", "Amazon");
book.getProperties().put("price", ".95");

entityManager.persist(book);

Hibernate executes the following SQL INSERT statement:

Hibernate 执行以下 SQL INSERT 语句:

INSERT INTO book (isbn, properties, id)
VALUES (
    '978-9730228236',
    '"author"=>"Vlad Mihalcea",
     "price"=>".95", "publisher"=>"Amazon",
     "title"=>"High-Performance Java Persistence"',
    1
)

And, when we fetch the Bookentity, we can see that all properties are fetched properly:

而且,当我们获取Book实体时,我们可以看到所有属性都被正确获取:

Book book = entityManager
.unwrap(Session.class)
.bySimpleNaturalId(Book.class)
.load("978-9730228236");

assertEquals(
    "High-Performance Java Persistence",
    book.getProperties().get("title")
);

assertEquals(
    "Vlad Mihalcea",
    book.getProperties().get("author")
);

For more details, check out this article.

有关更多详细信息,请查看这篇文章