点击刷新
刚从springmvc换到springboot，又遇到这个经典的问题
翻了半天帖子，很多傻逼的解决办法就是改为急加载
哦，好办法，头疼就把头砍了去呗，反正也不知道当初设置为懒加载是什么原因，也不关心急加载会有什么影响，这能是解决问题的态度吗
我又看了另外一些人的解决办法
在数据源处配置一下
spring.jpa.properties.hibernate.enable_lazy_load_no_trans =true
暂不清楚负面作用是什么，抽时间可以测一下，是不是破坏了连接池规则之类的
知道的朋友可以留言
第一条语句
select article0_.id as id1_0_0_, article0_.article_body_id as article_6_0_0_, article0_.body as body2_0_0_, article0_.publish_dt as publish_3_0_0_, article0_.title as title4_0_0_, article0_.writer as writer5_0_0_, articlebod1_.id as id1_1_1_, articlebod1_.body as body2_1_1_ from article article0_ left outer join article_body articlebod1_ on article0_.article_body_id=articlebod1_.id where article0_.id=?
第二条语句
select modifyreco0_.article_id as article_3_2_0_, modifyreco0_.id as id1_2_0_, modifyreco0_.id as id1_2_1_, modifyreco0_.publish_dt as publish_2_2_1_ from modify_record modifyreco0_ where modifyreco0_.article_id=? order by modifyreco0_.id asc
第三条语句（保存指令）
select article0_.id as id1_0_2_, article0_.article_body_id as article_6_0_2_, article0_.body as body2_0_2_, article0_.publish_dt as publish_3_0_2_, article0_.title as title4_0_2_, article0_.writer as writer5_0_2_, articlebod1_.id as id1_1_0_, articlebod1_.body as body2_1_0_, modifyreco2_.article_id as article_3_2_4_, modifyreco2_.id as id1_2_4_, modifyreco2_.id as id1_2_1_, modifyreco2_.publish_dt as publish_2_2_1_ from article article0_ left outer join article_body articlebod1_ on article0_.article_body_id=articlebod1_.id left outer join modify_record modifyreco2_ on article0_.id=modifyreco2_.article_id where article0_.id=? order by modifyreco2_.id asc
insert into modify_record (publish_dt) values (?)
update article set article_body_id=?, body=?, publish_dt=?, title=?, writer=? where id=?
update modify_record set article_id=? where id=?
如果用急加载
只发出了一条查询
select article0_.id as id1_0_0_, article0_.article_body_id as article_6_0_0_, article0_.body as body2_0_0_, article0_.publish_dt as publish_3_0_0_, article0_.title as title4_0_0_, article0_.writer as writer5_0_0_, articlebod1_.id as id1_1_1_, articlebod1_.body as body2_1_1_, modifyreco2_.article_id as article_3_2_2_, modifyreco2_.id as id1_2_2_, modifyreco2_.id as id1_2_3_, modifyreco2_.publish_dt as publish_2_2_3_ from article article0_ left outer join article_body articlebod1_ on article0_.article_body_id=articlebod1_.id left outer join modify_record modifyreco2_ on article0_.id=modifyreco2_.article_id where article0_.id=? order by modifyreco2_.id asc
然后这一句就不会在发出查询请求了，毕竟刚才已经查出来了
然后保存的时候
select article0_.id as id1_0_2_, article0_.article_body_id as article_6_0_2_, article0_.body as body2_0_2_, article0_.publish_dt as publish_3_0_2_, article0_.title as title4_0_2_, article0_.writer as writer5_0_2_, articlebod1_.id as id1_1_0_, articlebod1_.body as body2_1_0_, modifyreco2_.article_id as article_3_2_4_, modifyreco2_.id as id1_2_4_, modifyreco2_.id as id1_2_1_, modifyreco2_.publish_dt as publish_2_2_1_ from article article0_ left outer join article_body articlebod1_ on article0_.article_body_id=articlebod1_.id left outer join modify_record modifyreco2_ on article0_.id=modifyreco2_.article_id where article0_.id=? order by modifyreco2_.id asc
insert into modify_record (publish_dt) values (?)
update article set article_body_id=?, body=?, publish_dt=?, title=?, writer=? where id=?
update modify_record set article_id=? where id=?
四条语句
所以看起来
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