JPA和JSON运算符本机查询 [英] JPA and JSON operator native query

问题描述

我正在尝试使此查询在JPA中工作：

I'm trying to make this query work in JPA:

SELECT * FROM contrat WHERE contrat_json @> '{"nom" :"hever"}';

它与 postgresql 完美配合，但是当我将它与JPA集成，我得到以下错误：

It works perfectly with postgresql but when I integrate it with JPA, I get the following error:

该位置[1]的参数不存在

Parameter with that position [1] did not exist

我的代码：

 @Transactional
 @Query(nativeQuery = true,value = "select p from Contrat p where contrat_json @> '{\"nom\":\":nom\"}'")
    public List<Contrat> findByNomRestrict(@Param("nom") String nom);

我认为它无法识别 @> 尽管有本机查询，你有什么想法吗？

I think it does not recognize @> despite native query, do you have an idea?

推荐答案

使用PostgreSQL和JSON你可能会遇到需要？或其他奇怪的运算符，所以最好只使用它们的函数等价物。您可以在 psql 控制台中查找它们，例如 \ doS + @> 。

With PostgreSQL and JSON you'll probably run into needing ? or other strange operators, so it's better you just use their function equivalents, instead. You can look them up in the psql console like this \doS+ @>.

您的查询不是原生的，如参数所示。

Your query is not native, as the parameter says.

select p from Contrat p where...

只有在到达数据库时才会出错。

Will only give you an error when it reaches the database.

尝试类似

@Query(nativeQuery = true, value = "select * from Contrat where jsonb_contains(contrat_json, :nom )")

只需绑定{ \nom \：\+ param +\}作为参数

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