确定hibernate中查询列的数据类型的问题 [英] problem with determining data type of query column in hibernate

问题描述

我在hibernate中写下了下面的查询，它给了我NullPointerException

  SELECT new com.reddy.ReddyDTO（col1， IF（STRCMP（COL8，'zero'）= 0，'string'，col2））
 FROM entity 
  

然而，它抛出了NPE，因为它无法确定第二列的数据类型为字符串类型。

我可以看到它无法确定通过在 org.hibernate.hql.ast.tree.ConstructorNode.resolveConstructor（String path）方法中查看下面位置
的类型，字段 constructorArgumentTypes 具有值（org.hibernate.type.StringType，null）

<如何让hibernate理解这个字段是String数据类型？
（我已经试图把string放在双引号中，但是这给了我错误）

解决方案

嗯...这是什么 IF 表达式？我在参考文档中找不到它。据我所知，最接近的是一个 case 表达式：

简单案件，案件......当......然后......其他......结束，搜查案件，案件......然后......其他......结束

blockquote>

但是我怀疑你可以在构造函数表达式中使用它。
$ b 更新：看起来Hibernate比我想的要好，你可以在构造函数表达式中使用一个case：

  SELECT new com.reddy.ReddyDTO（col1，case col8 when'zero'then'string'else col2 end）FROM entity 
  

参考

• Hibernate核心参考指南
  
  
  
  
  • 14.10。 Expressions
  
  

I have written below query in hibernate and it is giving me NullPointerException

SELECT new com.reddy.ReddyDTO( col1, IF(STRCMP(COL8, 'zero')=0, 'string', col2) ) 
FROM entity

However it is throwing me NPE as it was unable to determine data type for second column as String type.

I can see that it was unable to determine a type by looking at below location in org.hibernate.hql.ast.tree.ConstructorNode.resolveConstructor(String path) method, the field constructorArgumentTypes is having values (org.hibernate.type.StringType, null)

How to make hibernate understand this column is of String data type? (I have already trying to put "string" in double-quotes, but that is giving me error)

解决方案

Hmm... What is this IF expression? I can't find it in the reference documentation. To my knowledge, the closest would be a case expression:

"Simple" case, case ... when ... then ... else ... end, and "searched" case, case when ... then ... else ... end

But I doubt you can use this in a constructor expression.

Update: It appears Hibernate is better than I thought, you can use a case in a constructor expression:

SELECT new com.reddy.ReddyDTO(col1, case col8 when 'zero' then 'string' else col2 end) FROM entity

Reference

• Hibernate Core Reference Guide
  • 14.10. Expressions

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