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jpa - java.lang.IllegalArgumentException : You have attempted to set a parameter value using a name of string that does not exist in the query string 异常

转载 作者:行者123 更新时间:2023-12-01 09:58:46 32 4
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我有以下命名查询:

@NamedQuery(
name = "Userdetails.findByUsername",
query = "SELECT u FROM Userdetails u WHERE u.username = :username"
)

当我尝试按如下方式执行时:

getEntityManager()
.createNamedQuery("Userdetails.findByUsername")
.setParameter("string", "%" + string + "%")
.getResultList();

然后我得到以下异常:

java.lang.IllegalArgumentException: You have attempted to set a parameter value using a name of string that does not exist in the query string SELECT u FROM Userdetails u WHERE u.username = :username.

这是怎么引起的,我该如何解决?

最佳答案

在您的 Userdetails.findByUsername 命名查询中,您有一个名为 username 的命名参数:

SELECT u FROM Userdetails u WHERE u.username = :username

然后您使用实体管理器调用它,但设置一个string 参数:

getEntityManager()
.createNamedQuery("Userdetails.findByUsername")
.setParameter("string", "%" + string + "%") //<-- check here
.getResultList();

用户名替换这个string参数:

getEntityManager()
.createNamedQuery("Userdetails.findByUsername")
.setParameter("username", "%" + string + "%") //<-- check here
.getResultList();

关于jpa - java.lang.IllegalArgumentException : You have attempted to set a parameter value using a name of string that does not exist in the query string 异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20080865/

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