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javaspring
Clay Banks
Clay Banks

Reputation: 4591

Spring - Parameter value did not match expected type [java.lang.Long]

I have the following line in my @Controller to 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

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

@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"):

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?

Upvotes: 5

Views: 22656

Answers (1)

Zbynek Vyskovsky - kvr000
Zbynek Vyskovsky - kvr000

Reputation: 18865

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.

Upvotes: 6

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