从JPA调用oracle函数 [英] Calling an oracle function from JPA

问题描述

我试图从JPA类中调用一个简单的函数，它根据一些计算返回一个数字并具有以下定义。

I am trying to call a simple function from a JPA class that returns a number based on some calculations and has the following definition.

'CREATE OR REPLACE FUNCTION CFB.FC_AMOUNT_CHECK(accountNumber IN VARCHAR2)
return NUMBER IS .....'

我试图通过以下方式从JPA调用此函数。

I am trying to call this function from JPA the following way.

StringBuilder sql = new StringBuilder("call CFB.FC_AMOUNT_CHECK(:accountNumber)");
Query query = em.createNativeQuery(sql.toString());
query.setParameter(1, '1234');
List<?> result = query.getResultList();

....

但是，当我执行这个类时，我总是得到以下异常：

However, when I execute this class, I get the below exception all the time:

java.lang.IllegalArgumentException: org.hibernate.QueryParameterException: could not locate named parameter [1]

我似乎无法得到JPA如何找不到参数1。 ...在过去的4个小时里，我一直在打破这个问题。任何人都可以建议如何得到我想要的结果？

I cant seem to get how JPA cannot find parameter 1....I have been breaking my head with this for the last 4 hours. Can anyone please suggest how to get the result I want?

推荐答案

如果你想在JPA中调用一个函数选择查询然后按照以下链接，它的工作原理：

If you are looking to call a function in JPA select query then follow below link, it works:

http://www.eclipse.org/eclipselink/documentation/2.5/jpa/extensions/j_func.htm

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