Spring - 参数值与预期类型不匹配 [java.lang.Long]
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Spring - Parameter value did not match expected type [java.lang.Long]
提问by Kurai Bankusu
I have the following line in my @Controllerto return results based on a string or partial string passed. The string may be numerical, in which case I parse it to a long value
我在我的以下行中@Controller根据传递的字符串或部分字符串返回结果。字符串可能是数字，在这种情况下，我将其解析为长值
public @ResponseBody Page<Object> searchUsers(
@RequestParam(value = "search-string", required = false) String string) {
List<User> results = myRepo.findByUId(Long.valueOf(searchString).longValue();
//do stuff
}
My Repo Method
我的回购方法
@Query("SELECT u FROM User b WHERE u.uid LIKE :id%")
List<User> findByIdStartsWith(@Param("uid") Long uid);
However I get the following Exception (say my parameter is "1"):
但是我得到以下异常（假设我的参数是“1”）：
java.lang.IllegalArgumentException: Parameter value [1%] did not match expected type [java.lang.Long (n/a)]
I don't think that I'm handling this parse correctly. Any ideas?
我认为我没有正确处理这个解析。有任何想法吗？
回答by Zbynek Vyskovsky - kvr000
Hibernate is trying casting the value "1%" to target type (User.id) which is Long. This is obviously impossible. LIKE works only with strings in hibernate, you may wrap with str(u.id) but see below.
Hibernate 正在尝试将值“1%”转换为 Long 的目标类型 (User.id)。这显然是不可能的。LIKE 仅适用于休眠中的字符串，您可以使用 str(u.id) 包装，但请参见下文。
