我将这个问题从前面的问题中分离出来，以消除“保存前选择”的混淆。在这个例子中，我试图通过主键执行一个简单的findOne（）。这是针对现有的sqlserver数据库，具有最新版本的Spring boot和Spring数据。

我有日志记录集，所以我可以看到hibernate生成的sql。在这个例子中，根据日志记录时间，这个查询大约需要4秒。这是使用主键查询。当我运行hibernate在db工具（如dbVisual alizer）中生成的sql时，它会在亚秒内返回，正如我所期望的那样。

我将hibernate打包日志记录增加到TRACE级别，尝试查看延迟在哪里，并在4秒延迟之前和之后发现以下内容：

2017-08-14 09:51:35.345 DEBUG 7532 --- [nio-8085-exec-1] org.hibernate.SQL                        : select customer0_.customer_id as customer1_4_0_, customer0_.first_name as first_na2_4_0_, customer0_.last_name as last_nam3_4_0_ from xbr_customer_tab customer0_ where customer0_.customer_id=?
Hibernate: select customer0_.customer_id as customer1_4_0_, customer0_.first_name as first_na2_4_0_, customer0_.last_name as last_nam3_4_0_ from xbr_customer_tab customer0_ where customer0_.customer_id=?
2017-08-14 09:51:35.470 TRACE 7532 --- [nio-8085-exec-1] o.h.r.j.i.ResourceRegistryStandardImpl   : Registering statement [org.apache.tomcat.jdbc.pool.StatementFacade$StatementProxy[Proxy=25287222; Query=select customer0_.customer_id as customer1_4_0_, customer0_.first_name as first_na2_4_0_, customer0_.last_name as last_nam3_4_0_ from xbr_customer_tab customer0_ where customer0_.customer_id=?; Delegate=SQLServerPreparedStatement:6]]
2017-08-14 09:51:35.471 TRACE 7532 --- [nio-8085-exec-1] o.h.e.jdbc.internal.JdbcCoordinatorImpl  : Registering last query statement [org.apache.tomcat.jdbc.pool.StatementFacade$StatementProxy[Proxy=25287222; Query=select customer0_.customer_id as customer1_4_0_, customer0_.first_name as first_na2_4_0_, customer0_.last_name as last_nam3_4_0_ from xbr_customer_tab customer0_ where customer0_.customer_id=?; Delegate=SQLServerPreparedStatement:6]]
2017-08-14 09:51:35.479 TRACE 7532 --- [nio-8085-exec-1] o.h.type.descriptor.sql.BasicBinder      : binding parameter [1] as [VARCHAR] - [40666316]
2017-08-14 09:51:35.488 TRACE 7532 --- [nio-8085-exec-1] o.h.l.p.e.i.AbstractLoadPlanBasedLoader  : Bound [2] parameters total
2017-08-14 09:51:39.426 TRACE 7532 --- [nio-8085-exec-1] o.h.r.j.i.ResourceRegistryStandardImpl   : Registering result set [SQLServerResultSet:6]
2017-08-14 09:51:39.434 TRACE 7532 --- [nio-8085-exec-1] o.h.l.p.e.p.i.ResultSetProcessorImpl     : Processing result set
2017-08-14 09:51:39.434 DEBUG 7532 --- [nio-8085-exec-1] o.h.l.p.e.p.i.ResultSetProcessorImpl     : Starting ResultSet row #0
2017-08-14 09:51:39.436 DEBUG 7532 --- [nio-8085-exec-1] l.p.e.p.i.EntityReferenceInitializerImpl : On call to EntityIdentifierReaderImpl#resolve, EntityKey was already known; should only happen on root returns with an optional identifier specified
2017-08-14 09:51:39.436 TRACE 7532 --- [nio-8085-exec-1] l.p.e.p.i.EntityReferenceInitializerImpl : hydrating entity state

我也想知道为什么它说2个参数绑定，而sql中只有1个。

知道为什么这个选择需要这么长时间吗？尤其是在我的Spring应用程序中，而不是在另一个客户端（如dbVisualizer）中？

这是实体：

import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "my_customer_table")
public class Customer {

    @Id
    private String customer_id;
    private String first_name;
    private String last_name;

    protected Customer() {}


    public Customer(String firstName, String lastName) {
        this.first_name = firstName;
        this.last_name = lastName;
    }

}

这是CustomerRepository

import com.....Customer;
import org.springframework.data.repository.CrudRepository;

public interface CustomerRepository extends CrudRepository<Customer, String> {
}

以及查找客户的代码，来自@Service类，其中CustomerRepository是@Autow的：

Customer customer = customerRepository.findOne(customerId);

这是生成的sql：

select customer0_.customer_id as customer1_4_0_, customer0_.first_name as first_na2_4_0_, customer0_.last_name as last_nam3_4_0_ from my_customer_table customer0_ where customer0_.customer_id=?