Rest API order by name - rest

Question

I'm trying to create my own REST API and I'm having trouble trying to order my data by name. currently, I am able to display all the data from the styles table, however, I wish to sort them alphabetically.
I was able to do a filter by extracting the year from the date and checking if that was in the database, this is shown in
public List<Beers> getAllBeersByYear(int year) {
EntityManager em = DBUtil.getEMF().createEntityManager();
List<Beers> list = null;
List<Beers> beersToRemove = new ArrayList<>();
try {
list = em.createNamedQuery("Beers.findAll", Beers.class)
.getResultList();
if (list == null || list.isEmpty()) {
list = null;
}
} finally {
em.close();
}
Calendar cal = Calendar.getInstance();
for (Beers beer : list) {
cal.setTime(beer.getLastMod());
if (cal.get(Calendar.YEAR) != year) {
beersToRemove.add(beer);
}
}
list.removeAll(beersToRemove);
return list;
}
the controller is
#GetMapping(produces=MediaType.APPLICATION_JSON_VALUE)
public List<Styles> GetAllStyles() {
return service.getAllStyles();
}
would it be possible to do something similar to the service and controller where instead of filtering the data, it can sort by the name of a column
the service is
public List<Styles> getAllStyles() {
EntityManager em = DBUtil.getEMF().createEntityManager();
List<Styles> list = null;
try {
list = em.createNamedQuery("Styles.findAll", Styles.class)
.getResultList();
if (list == null || list.isEmpty()) {
list = null;
}
} finally {
em.close();
}
return list;
}
the JPA I am using is
#Entity
#Table(name = "styles")
#NamedQueries({
#NamedQuery(name = "Styles.findAll", query = "SELECT s FROM Styles s"),
#NamedQuery(name = "Styles.findById", query = "SELECT s FROM Styles s WHERE s.id = :id"),
#NamedQuery(name = "Styles.findByCatId", query = "SELECT s FROM Styles s WHERE s.catId = :catId"),
#NamedQuery(name = "Styles.findByStyleName", query = "SELECT s FROM Styles s WHERE s.styleName = :styleName"),
#NamedQuery(name = "Styles.findByLastMod", query = "SELECT s FROM Styles s WHERE s.lastMod = :lastMod")})
public class Styles implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Basic(optional = false)
#Column(name = "id")
private Integer id;
#Basic(optional = false)
#NotNull
#Column(name = "cat_id")
private int catId;
#Basic(optional = false)
#NotNull
#Size(min = 1, max = 255)
#Column(name = "style_name")
private String styleName;
#Basic(optional = false)
#NotNull
#Column(name = "last_mod")
#Temporal(TemporalType.TIMESTAMP)
private Date lastMod;
public Styles() {
}
public Styles(Integer id) {
this.id = id;
}
public Styles(Integer id, int catId, String styleName, Date lastMod) {
this.id = id;
this.catId = catId;
this.styleName = styleName;
this.lastMod = lastMod;
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public int getCatId() {
return catId;
}
public void setCatId(int catId) {
this.catId = catId;
}
public String getStyleName() {
return styleName;
}
public void setStyleName(String styleName) {
this.styleName = styleName;
}
public Date getLastMod() {
return lastMod;
}
public void setLastMod(Date lastMod) {
this.lastMod = lastMod;
}
#Override
public int hashCode() {
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
}
#Override
public boolean equals(Object object) {
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof Styles)) {
return false;
}
Styles other = (Styles) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
return false;
}
return true;
}
#Override
public String toString() {
return "Service.Styles[ id=" + id + " ]";
}
}

Answer 1

You can use #OrderBy annotation
https://www.logicbig.com/tutorials/java-ee-tutorial/jpa/order-by-annotation.html

Answer 2

Just create another query:
#NamedQuery(name = "Styles.findAll", query = "SELECT s FROM Styles s ORDER BY s.name")
And why are you filtering in the code when you can add a query with a where condition?