在Hibernate中MapPostgreSQL LTREE列时出错

noj0wjuj  于 2023-03-19  发布在  PostgreSQL
关注(0)|答案(4)|浏览(127)

我尝试在Hibernate中Mappostgresql ltree列,如下所示:

所在实体:

private String path;

@Column(name="org_path", columnDefinition="ltree")
public String getPath() {
   return path;

表格结构:

CREATE TABLE relationship (
    relationship_id int4 NOT NULL,
    parent_organization_id uuid NOT NULL,
    child_organization_id uuid NOT NULL,
    org_path ltree NOT NULL,
    CONSTRAINT relationship_pk PRIMARY KEY (relationship_id),
    CONSTRAINT organization_fk3 FOREIGN KEY (parent_organization_id) REFERENCES organization(organization_id) ON DELETE RESTRICT ON UPDATE RESTRICT,
    CONSTRAINT organization_fk4 FOREIGN KEY (child_organization_id) REFERENCES  organization(organization_id) ON DELETE RESTRICT ON UPDATE RESTRICT
)

出现以下错误:

wrong column type encountered in column [org_path] in table [relationship]; found [“schemaName"."ltree" (Types#OTHER)], but expecting [ltree (Types#VARCHAR)]

有人能帮助解决这个问题吗?

0pizxfdo

0pizxfdo1#

在Java中实现一个自定义LTreeType类,如下所示:

public class LTreeType implements UserType {

    @Override
    public int[] sqlTypes() {
        return  new int[] {Types.OTHER};
    }

    @SuppressWarnings("rawtypes")
    @Override
    public Class returnedClass() {
        return String.class;
    }

    @Override
    public boolean equals(Object x, Object y) throws HibernateException {
        return x.equals(y);
    }

    @Override
    public int hashCode(Object x) throws HibernateException {
        return x.hashCode();
    }

    @Override
    public Object nullSafeGet(ResultSet rs, String[] names, Object owner)
            throws HibernateException, SQLException {
        return rs.getString(names[0]);
    }

    @Override
    public void nullSafeSet(PreparedStatement st, Object value, int index)
            throws HibernateException, SQLException {
        st.setObject(index, value, Types.OTHER);
    }

    @Override
    public Object deepCopy(Object value) throws HibernateException {
        return new String((String)value);
    }

    @Override
    public boolean isMutable() {
        return false;
    }

    @Override
    public Serializable disassemble(Object value) throws HibernateException {
        return (Serializable)value;
    }

    @Override
    public Object assemble(Serializable cached, Object owner)
            throws HibernateException {
        return cached;
    }

    @Override
    public Object replace(Object original, Object target, Object owner)
            throws HibernateException {
        // TODO Auto-generated method stub
        return deepCopy(original);
    }

}

并对Entity类进行如下注解:

@Column(name = "path", nullable = false, columnDefinition = "ltree")
    @Type(type = "LTreeType")
    private String path;
5m1hhzi4

5m1hhzi42#

直到我也创建了一个LQueryType,就像为LTreeType提供的类@ arnabbis一样,我的代码只知道字符串,但Postgres不知道如何使用字符串的ltree。

ltree ~ lquery
ltree @> ltree

我的KotlinJPA是这样的:

val descendantIds = treeRepo.findAllDescendantIds("*.$id.*{1,}")
. . .
@Query(
    "SELECT node_id FROM tree WHERE path ~ CAST(:idQuery AS lquery);"
    , nativeQuery = true)
fun findAllDescendantIds(@Param("idQuery") idQuery: String): Array<Long>
dba5bblo

dba5bblo3#

只需在@anarbbswas代码上添加此修改,它就会正常工作

@Override
public Object nullSafeGet(ResultSet rs, String[] names,SharedSessionContractImplementor session, Object owner)
        throws HibernateException, SQLException {
    return rs.getString(names[0]);
}

@Override
public void nullSafeSet(PreparedStatement st, Object value, int index, SharedSessionContractImplementor session) throws HibernateException, SQLException {
    st.setObject(index, value, Types.OTHER);
}

@Override
public Object deepCopy(Object value) throws HibernateException {
    if (value == null)
        return null;
    if (! (value instanceof String))
        throw new IllegalStateException("Expected String, but got: " + value.getClass());
    return value;
}
jljoyd4f

jljoyd4f4#

在Hibernate 6.1.7.Final中,这不再起作用了。有关于如何迁移的建议吗?

@Column(name = "path", nullable = false, columnDefinition = "ltree")
@Type(type = "LTreeType")
private String path;

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