Question

框架：Spring 4.0.7和Hibernate 4.3.6

我们遇到问题PagingAndSortingRepository @Query需要花费大量时间才能完成，因为它似乎在考虑页数据限制之前正在运行记录集数据和构建结构。

在非联合方案中，fetch first n rows按预期显示，结果集小得多，使得第一批页面上的选择非常活泼。

我们遇到的最初问题是我们无法对详细记录执行@Query inner join fetch（强制加入）。删除fetch显然让我们离开了地面，但创建了一个嵌套的SQL循环，这是不可取的。

为了解决这个问题，我们添加了countQuery，它让我们得到了正确的计数，但现在看来在返回分页结果之前将整个结构加载到内存中。打开调试会显示正在加载的记录。fetch first n rows尚未添加。

在循环中运行嵌套的SQL不是我们的选项，因此删除fetch无济于事。

@Query(value = "select e from InvoiceHistoryHeader e inner join fetch e.details f where e.company = ?1 and e.division = ?2 and e.customerNumber = ?3)", countQuery = "select count(e) from InvoiceHistoryHeader e where e.company = ?1 and e.division = ?2 and e.customerNumber = ?3)")
Page<InvoiceHistoryHeader> findByCustomerNumber(String company, String division, String customerNumber, Pageable pageable);

任何人都可以提供此问题的解决方案，或提供可能有用的解决方法吗？我可以根据需要提供更多细节。

编辑（2014年11月11日）

休眠输出（无提取）：

Hibernate: select count(trim(invoicehis0_.HH_TID)) as col_0_0_ from invhh invoicehis0_ where trim(invoicehis0_.HH_CO)=? and trim(invoicehis0_.HH_DIV)=? and trim(invoicehis0_.HH_CUS)=? and trim(invoicehis0_.HH_BR)=?
Hibernate: select invoicehis0_.hh_tid as hh_tid1_7_, trim(invoicehis0_.HH_CO) as hh_co2_7_, trim(invoicehis0_.HH_CUS) as hh_cus3_7_, trim(invoicehis0_.HH_DTI) as hh_dti4_7_, trim(invoicehis0_.HH_DIV) as hh_div5_7_, trim(invoicehis0_.HH_ORD) as hh_ord6_7_, trim(invoicehis0_.HH_BR) as hh_br7_7_ from invhh invoicehis0_ where trim(invoicehis0_.HH_CO)=? and trim(invoicehis0_.HH_DIV)=? and trim(invoicehis0_.HH_CUS)=? and trim(invoicehis0_.HH_BR)=? fetch first 20 rows only 
Hibernate: select details0_.hd_tid as hd_tid2_7_0_, trim(details0_.HD_SEQ) as hd_seq1_6_0_, trim(details0_.HD_TID) as hd_tid2_6_0_, details0_.hd_seq as hd_seq1_6_1_, details0_.hd_tid as hd_tid2_6_1_, trim(details0_.HD_DTL) as hd_dtl3_6_1_, trim(details0_.HD_LIN) as hd_lin4_6_1_ from invhd details0_ where details0_.hd_tid=?
Hibernate: select details0_.hd_tid as hd_tid2_7_0_, trim(details0_.HD_SEQ) as hd_seq1_6_0_, trim(details0_.HD_TID) as hd_tid2_6_0_, details0_.hd_seq as hd_seq1_6_1_, details0_.hd_tid as hd_tid2_6_1_, trim(details0_.HD_DTL) as hd_dtl3_6_1_, trim(details0_.HD_LIN) as hd_lin4_6_1_ from invhd details0_ where details0_.hd_tid=?
<above two SQL repeated 18 more times>

Hibernate输出（fetch和countQuery）：

Hibernate: select invoicehis0_.hh_tid as hh_tid1_7_0_, details1_.hd_seq as hd_seq1_6_1_, details1_.hd_tid as hd_tid2_6_1_, trim(invoicehis0_.HH_CO) as hh_co2_7_0_, trim(invoicehis0_.HH_CUS) as hh_cus3_7_0_, trim(invoicehis0_.HH_DTI) as hh_dti4_7_0_, trim(invoicehis0_.HH_DIV) as hh_div5_7_0_, trim(invoicehis0_.HH_ORD) as hh_ord6_7_0_, trim(invoicehis0_.HH_BR) as hh_br7_7_0_, trim(details1_.HD_DTL) as hd_dtl3_6_1_, trim(details1_.HD_LIN) as hd_lin4_6_1_, details1_.hd_tid as hd_tid2_7_0__, trim(details1_.HD_SEQ) as hd_seq1_6_0__, trim(details1_.HD_TID) as hd_tid2_6_0__ from invhh invoicehis0_ inner join invhd details1_ on trim(invoicehis0_.HH_TID)=details1_.hd_tid where trim(invoicehis0_.HH_CO)=? and trim(invoicehis0_.HH_DIV)=? and trim(invoicehis0_.HH_CUS)=? and trim(invoicehis0_.HH_BR)=?

Answer 1

在连接和分页中使用fetch时，Hibernate执行内存分页，这意味着它会加载内存中的所有记录。不要在分页中使用fetch join。

如何在PagingAndSortingRepository @Query中限制结果？

1 个答案: