将MySQL查询转换为Hibernate Criteria [英] Converting MySQL query to Hibernate Criteria

问题描述

  SELECT * FROM test.pchi new INNER JOIN rlhi old ON new_id = old.menu_id where new.name ='？'
  

类似于：

 从db.employee emp中选择* INNER JOIN db.table on emp_tableID = table.id where emp.name ='？'
  

如果你可以告诉我如何做一个投影，这将是很棒的......如下：

 选择emp.name，emp。 sex，table.brand from .... 
  

我尝试使用抓取，但我很新对此并不断发生一些奇怪的错误。有人可以请示范如何做到这一点？

如何？

  sess.createCriteria（pchi.class ）/ ** / 
 .setFetchMode（rlhi，FetchMode.JOIN）
 .add（Restrictions.eq（new_id，rlhi.menu_id））
 .add（Restrictions .eq（name，SOME INPUT））
  

解决方案

  sess.createCriteria（Pchi.class）
 .setFetchMode（rlhi，FetchMode.JOIN）//注意rlh1是Pchi类中的属性名称
 .add（Restrictions.eq（name，SOME INPUT））; 
  

在你的课堂里你有类似的东西

  class Pchi {
 @ManyToOne（fetch = FetchType.LAZY）
 @JoinColumn（name =new_id，referencedColumnName =menu_id）
私人Rlhi rlhi; 
} 
 
 class Rlhi {
 @OneToMany（mappedBy =rlhi）
 private< Set> Pchi pchis; 
} 
  

注当您使用 SET 应该覆盖 equals（）和 hashCode（）方法正常工作

SELECT * FROM test.pchi new INNER JOIN rlhi old ON new_id = old.menu_id where new.name='?'

Similar to:

Select * from db.employee emp INNER JOIN db.table on emp_tableID = table.id where emp.name = '?'

If you could tell me how to do a projection, that would be awesome... as in:

Select emp.name, emp.sex, table.brand from ....

I tried using fetch, but I'm very new to this and keep getting some weird errors. Could someone please demonstrate how to do this?

How about this?

sess.createCriteria(pchi.class)/**/
              .setFetchMode("rlhi", FetchMode.JOIN)
              .add(Restrictions.eq("new_id", "rlhi.menu_id"))
              .add(Restrictions.eq("name", "SOME INPUT"))

解决方案

sess.createCriteria(Pchi.class)
              .setFetchMode("rlhi", FetchMode.JOIN) //note that rlh1 is the property name in Pchi class
              .add(Restrictions.eq("name", "SOME INPUT"));

In your class you have something like this

class Pchi{
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name="new_id", referencedColumnName="menu_id")
private Rlhi rlhi;
}

class Rlhi{
@OneToMany(mappedBy="rlhi")
private <Set> Pchi pchis;
}

NOTE When you use SET should override equals() and hashCode() method to work properly

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