Spring MVC - Search using Param - database

Question

I want to search something from database example i have three textfields
First Name , Last Name and Gender
if i input there
Some Firstname , Some Lastname , Male I want to search all the people with that certain name and gender and display it how do I do this?
Note: I am just new to Spring and just started learning thanks
Controller:
#RequestMapping(value = "/student/search", method = RequestMethod.GET)
public String searchStudent(#Param("name") String name, #Param("type") String type, Map<String, Object> model) {
Map<String, String> params = new HashMap<String, String>();
params.put("name", name);
params.put("type", gender);
model.put("students",studentRepository.findAll(StudentSpecification.search(params)));
return "/students/list";
}
}
Specification:
public static Specification<Student> search(final Map<String, String> params) {
return (root, query, criteriaBuilder) -> {
List<Predicate> predicates = new ArrayList<Predicate>();
params.forEach((k, v) -> {
if (StringUtils.equals(k, "name")) {
if (StringUtils.isNotBlank(v)) {
Predicate hasFirstName =
criteriaBuilder.like(root.join("user").<String>get("firstName"), "%" + v + "%");
Predicate hasLastName =
criteriaBuilder.like(root.join("user").<String>get("lastName"), "%" + v + "%");
predicates.add(criteriaBuilder.or(hasFirstName, hasLastName));
}
}
});
return criteriaBuilder.and(predicates.toArray(new Predicate[] {}));
};
}
}

Answer 1

I presume from your code that you are using Spring data jpa, with specification, right?
if so, you need to generate MetaModel of your entities, add the code below if you are using maven, and this will do it for you
<plugin>
<groupId>org.bsc.maven</groupId>
<artifactId>maven-processor-plugin</artifactId>
<version>2.0.5</version>
<executions>
<execution>
<id>process</id>
<goals>
<goal>process</goal>
</goals>
<phase>generate-sources</phase>
<configuration>
<processors>
<processor>org.hibernate.jpamodelgen.JPAMetaModelEntityProcessor</processor>
</processors>
</configuration>
</execution>
</executions>
<dependencies>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-jpamodelgen</artifactId>
<version>4.3.6.Final</version>
</dependency>
</dependencies>
</plugin>
I suppose you have an entity called Student, this plugin declaration will generate a classe suffixed with underscore 'Student_'
and then you could implement the specification interface, something like this
public class StudentSpecification {
public static Specification<Student> search(final Map<String, String> params) {
return new Specification<Student>() {
#Override
public Predicate toPredicate(Root<Student> studentRoot, CriteriaQuery<?> query, CriteriaBuilder cb) {
Predicate res = null;
if(params.containKey("name"))
res = cb.equal(studentRoot.get(Student_.name), params.get("name"));
if(params.containKey("type"))
if(res ==null)
res = cb.equal(studentRoot.get(Student_.name), params.get("type"));
else
res = cb.and(res, cb.equal(studentRoot.get(Student_.name), params.get("type")));
return res;
}
}
}
and of course your StudentRepository class should extends JpaSpecificationExecutor, something like this
interface StudentRepository extends JpaSpecificationExecutor<Student> {
}

Answer 2

You can do this using #Query, add method in the repository:
#Query("SELECT e FROM STUDENT e WHERE e.name = ?1 AND e.type = ?2")
public List<student> findAll(String name, String type);
Use that method in the controller:
#RequestMapping(value = "/student/search", method = RequestMethod.GET)
public String searchStudent(#Param("name") String name, #Param("type") String type, Map<String, Object> model) {
model.put("students",studentRepository.findAll(name, type));
return "/students/list";
}