如何为JSONB列构建JPA谓词？

import javax.persistence.criteria.CriteriaBuilder;
import javax.persistence.criteria.CriteriaQuery;
import javax.persistence.criteria.Predicate;
import javax.persistence.criteria.Root;
import javax.persistence.criteria.Path;
import javax.persistence.criteria.Expression;
import org.hibernate.query.criteria.internal.path.SingularAttributePath;

// 假设有一个实体类名为Entity，包含一个名为jsonbColumn的JSONB列

public List<Entity> searchByJsonbColumn(String key, String value) {
    CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<Entity> query = cb.createQuery(Entity.class);
    Root<Entity> root = query.from(Entity.class);
    
    // 获取JSONB列的路径
    Path<Object> jsonbPath = root.get("jsonbColumn");
    
    // 构建JSONB列的键值对
    Expression<String> jsonbKey = cb.literal(key);
    Expression<String> jsonbValue = cb.literal(value);
    Expression<String> jsonbKeyValue = cb.function("jsonb_extract_path_text", String.class, jsonbPath, jsonbKey);
    
    // 构建谓词
    Predicate predicate = cb.equal(jsonbKeyValue, jsonbValue);
    
    // 执行查询
    query.select(root).where(predicate);
    return entityManager.createQuery(query).getResultList();
}