我有人,我只想得到以某个字母开头的字母(字母来自输入字段)。我有查询,但我不能'使用'它。 我怎样才能做到这一点?
存储库:
@Repository
public interface PersonRepository extends JpaRepository<Person, Integer>{
@Query("From Person where firstname like CONCAT(:firstname,'%')")
Stream<Person> findAllWithSearchParams(@Param("firstname") String firstname);
}
服务:
@Service
public class PersonService {
@Autowired
private PersonRepository personRepository;
public Stream<Person> all(Person mysearch){
return personRepository
.findAll(Example.of(mysearch))
.stream()
.map(Person::fromPerson);
}
}
班主任:
public class Person {
public Integer index;
public String firstname;
public String lastname;
@JsonFormat(pattern="dd.MM.yyyy")
public Date exdate;
public String insnr;
private Person(Integer index, String firstname, String lastname, Date exdate, String insnr){
this.index=index;
this.firstname=firstname;
this.lastname=lastname;
this.exdate=exdate;
this.insnr=insnr;
}
public static Person fromPerson(Person person){
return person == null ? null : new Person(person.getIndex(), person.getFirstname(), person.getLastname(), person.getExdate(), person.getInsnr());
}
}
控制器:
@Autowired
private PersonService personService;
@RequestMapping(value="/person/list/**")
public List<Person> loadPersonList(
@RequestParam(value = "firstname" ,required=false) String firstname) throws ParseException {
mysearch.setFirstname(firstname);
return personService.all(mysearch).collect(Collectors.toList());
}
答案 0 :(得分:0)
如果我理解正确,您需要一个查找以某个字母或单词开头的人的查询。所以,你只需要:
@Repository
public interface PersonRepository extends CrudRepository<Person, Integer> {
List<Person> findByFirstnameStartingWith(String firstname);
}