爲什麼一個簡單的Hibernate findOne（）通過主鍵佔用這麼久？

Question

我把這個從an earlier question分開，以消除'選擇之前保存'的困惑。在這個例子中，我試圖通過主鍵做一個簡單的findOne（）。這是針對現有的sqlserver數據庫，以及最新版本的spring引導和spring數據。

我有日誌記錄集，所以我可以看到休眠生成的SQL。在這個例子中，根據記錄時間，這個查詢大約需要4秒。這是使用主鍵查詢的。當我運行hibernate在像dbvisualizer這樣的db工具中生成的sql時，它會在我所期望的秒內返回。

我增加了休眠包裝記錄到跟蹤級別，要儘量看看那裏的延遲，並發現了下面，前後4秒後：

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 

我也想知道爲什麼它說2個參數綁定，當在sql中只有1。

任何想法，爲什麼這個選擇需要這麼久？特別是只在我的春天的應用程序，而不是像dbvisualizer的另一個客戶端？

這裏是實體：

：

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; 
    } 

} 

這裏的CustomerRepository

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

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

和代碼查找一個客戶，從@服務類，其中CustomerRepository中@Autowired

Customer customer = customerRepository.findOne(customerId); 

這裏是生成的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=? 

Answer 1

通過將以下參數添加到數據源中的連接字符串中進行修復。

sendStringParametersAsUnicode=false 

更多細節： http://www.jochenhebbrecht.be/site/2014-05-01/java/fixing-slow-queries-running-sql-server-using-jpa-hibernate-and-jtds