如何将 PostgreSQL hstore/json 与 JdbcTemplate 一起使用

postgresql java spring jdbctemplate hstore

2022-09-02 19:54:41

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

例如：

hstore：

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

对于 Json

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

答案 1

虽然对于答案（对于插入部分）来说已经很晚了，但我希望它可能对其他人有用：

取哈希图中的键/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\"}");

使用以下方法之一将它们插入PostgreSQL：

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)

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

3）在 POJO 中设置 hstoreMap/jsonbObj （map 类型的 hstoreCol 和 jsonbObjCol 是 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 );

并获取值：

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

答案 2

甚至比更简单，您可以使用休眠类型开源项目来持久保存 HStore 属性。JdbcTemplate

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>

然后，假设您具有以下实体：Book

@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
}

请注意，我们使用注释对实体属性进行了注释，并指定了以前通过定义的类型以使用PostgreSQLHStoreType自定义Hibernate Type。properties@Typehstore@TypeDef

现在，在存储以下实体时：Book

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", "$44.95");
 
entityManager.persist(book);

休眠执行以下 SQL INSERT 语句：

INSERT INTO book (isbn, properties, id)
VALUES (
    '978-9730228236',
    '"author"=>"Vlad Mihalcea",
     "price"=>"$44.95", "publisher"=>"Amazon",
     "title"=>"High-Performance Java Persistence"',
    1
)

而且，当我们获取实体时，我们可以看到所有属性都已正确获取：Book

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")
);