spring - 如何将迭代器值包装到简单的java类中

Question

我是 java 新手。我想将 result 的值包装在简单的 java 类中。

Iterator<Map<String,Object>> result=template.query(cypher,params);

任何帮助将不胜感激。

Answer 1

If you're using the template.query then you can either have it mapped to a domain entity or to the Map (and then you build the POJO yourself).

Otherwise, you can use a @Query in a repository and map it to a query result class. For example

 @Query("MATCH (user:User) WHERE user.gender={0} RETURN user.name AS UserName, user.gender AS UserGender, user.account as UserAccount, user.deposits as UserDeposits")
 Iterable<RichUserQueryResult> findUsersByGender(Gender gender);

@QueryResult
public class RichUserQueryResult {

    private Gender userGender; 
    private String userName;
    private BigInteger userAccount;
    private BigDecimal[] userDeposits;

    public Gender getUserGender() {
        return userGender;
    }

    public String getUserName() {
        return userName;
    }

    public BigInteger getUserAccount() {
        return userAccount;
    }

    public BigDecimal[] getUserDeposits() {
        return userDeposits;
    }
}