Java 如何在 JPA 中构建插入查询

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/43347602/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-12 00:51:31  来源:igfitidea点击:

How to construct an insert query in JPA

javahibernatejpaentity

提问by Armine

I am trying to insert data into a table having columns (NAME, VALUE)with

我试图将数据插入到具有表列(名称,值)

Query query = em.createQuery("INSERT INTO TestDataEntity (NAME, VALUE) VALUES (:name, :value)");
query.setParameter("name", name);
query.setParameter("value", value);
query.executeUpdate();

and getting the following exception:

并得到以下异常:

ERROR org.hibernate.hql.internal.ast.ErrorCounter - line 1:42: unexpected token: VALUES 

Also, I cannot insert a record using a native query either:

此外,我也无法使用本机查询插入记录:

Query query = em.createNativeQuery("INSERT INTO TEST_DATA (NAME, VALUE) VALUES (:name, :value);");
query.setParameter("name", name);
query.setParameter("value", value);
query.executeUpdate();

Another exception is being thrown:

正在抛出另一个异常:

javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute statement

The question is:

问题是:

  • What is wrong with the query string?
  • 查询字符串有什么问题?

Many thanks.

非常感谢。

采纳答案by Armine

I solved the issue.

我解决了这个问题。

According to this,

根据

There is no INSERT statement in JPA.

JPA 中没有 INSERT 语句。

But I could solve the issue with native query: I have mistakenly put a redundant ; at the end of the query, so the issue solved by removing it.

但是我可以用本机查询解决这个问题:我错误地放置了一个多余的 ; 在查询结束时,因此通过删除它解决了问题。

回答by Má?a - Stitod.cz

I found two examples where the author uses the insert in a native query (firstand second). Then, your query could be:

我发现了两个例子,作者在本机查询中使用插入(firstsecond)。然后,您的查询可能是:

Query query = em.createQuery("INSERT INTO TestDataEntity (NAME, VALUE) VALUES (?, ?)");
query.setParameter(1, name);
query.setParameter(2, value);
query.executeUpdate();

Try this.

尝试这个。

回答by Rajdeep Singh

I was able to do this using the below - I just had a Table name Bike with 2 fields id and name . I used the below to insert that into the database.

我能够使用下面的方法来做到这一点 - 我只有一个表名 Bike 有 2 个字段 id 和 name 。我使用以下内容将其插入到数据库中。

Query query = em.createNativeQuery("INSERT INTO Bike (id, name) VALUES (:id , :name);");
em.getTransaction().begin();
query.setParameter("id", "5");
query.setParameter("name", "Harley");
query.executeUpdate();
em.getTransaction().commit();