在 Spring 数据 JPA 中使用 @Query 进行动态查询？

Question

I am using Specifications in my Spring Boot app and can filter result by different filter options. However, I need to use special filter with @Query in my repository method and as far as I see, I cannot build a dynamic WHERE clause in this query.

There are also QueryDSL and CriteriaAPI options, but I cannot find an example for using them in @Query .

So, is it possible to dynamically build WHERE clause or create filter for the query in @Query ? Here is my method:

// there are more filters that omitted for brevity 

@Query("SELECT r FROM Recipe r WHERE r.title LIKE %:text%")
Page<Recipe> findByFields(@Param("text") String text);

I tried to use my specification in this method, but it is not possible to use them with @Query :((

Update:

@Entity
public class Recipe {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    private String title;

    @Enumerated(value = EnumType.STRING)
    private HealthLabel healthLabel;

    // code omitted for brevity

    @OneToMany(mappedBy = "recipe", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<RecipeIngredient> recipeIngredients = new ArrayList<>();
}

@Entity
public class RecipeIngredient {

    @EmbeddedId
    private RecipeIngredientId recipeIngredientId = new RecipeIngredientId();

    @Column(nullable = false)
    private BigDecimal amount;

    @ManyToOne(optional = true, fetch = FetchType.LAZY)
    @MapsId("recipeId")
    @JoinColumn(name = "recipe_id", referencedColumnName = "id")
    private Recipe recipe;

    @ManyToOne(optional = true, fetch = FetchType.LAZY)
    @MapsId("ingredientId")
    @JoinColumn(name = "ingredient_id", referencedColumnName = "id")
    private Ingredient ingredient;
}

@Entity
public class Ingredient {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(unique = true, nullable = false, length = 50)
    private String name;

    @OneToMany(mappedBy = "ingredient", cascade = CascadeType.ALL)
    private Set<RecipeIngredient> recipeIngredients = new HashSet<>();
}

Here is also my enum that I cannot filter by:

@Getter
@AllArgsConstructor
public enum HealthLabel {

    DEFAULT("Default"),
    EGG_FREE("Egg-free"),
    VEGETARIAN("Vegetarian"),
    WHEAT_FREE("Wheat-free");

    private String label;
}

Answer 1

@Query can only do static queries.

If you want something more dynamic you have to use another feature, for example Specifications or even fall back to custom method implementations .

Answer 2

You can try like this:

@Query("SELECT r FROM Recipe r WHERE r.title LIKE %?1%")
Page<Recipe> findByFields(String text);

Answer 3

There is nothing preventing you from using a Specification (which is basically the Criteria API but easier to use). You basically want the dynamic version of the following JPQL.

SELECT r FROM Recipe r 
JOIN r.recipeIngredients ri 
JOIN ri.ingredient i
WHERE i.name LIKE :name

Now if you can write it in JPQL you can also use the Criteria API to write it (generally speaking).

public static Specification<Recipe> findByIngredientName(String name) {
  return (root, query, criteriaBuilder) -> {
     Join<Recipe, RecipeIngredient> ingredients = root.join("ingredients");
     Join< RecipeIngredient, Ingredient> ingredient = ingredients.join("ingredient");      
     return criteriaBuilder.like(ingredient.get("name"), "%" + name + "%"); 
  };
}

That is the specification to retrieve a Recipe containing an Ingredient with a name like something. You need 2 joins no query just a join.

You can now even combine multiple predicates by using Predicate.and . So if you create another Predicate for the Recipe.name you can just chain them together, JPA will handle the rest.

public static Specification<Recipe> findName(String name) {
  return (root, query, criteriaBuilder) -> { 
     return criteriaBuilder.like(root.get("title"), "%" + name + "%"); 
  };
}

Specification<Recipe> specs = findByName("recipe").and(findByIngredientName("ingredient"));
recipeRepository.findAll(specs);

That is all you need. What and how you receive those parameters that is up to you how you build the request/response objects and is highly subjective. If you want a sort order use the findAll which takes a Specification and a Sort

Specification<Recipe> specs = findByName("recipe").and(findByIngredientName("ingredient"));
recipeRepository.findAll(specs, Sort.by("title"));

The above will generate a SQL in the form of

SELECT recipesapp0_.id as id1_1_, recipesapp0_.title as title2_1_ FROM recipes_application$recipe recipesapp0_ 
INNER JOIN recipes_application$recipe_ingredient recipeingr1_ ON recipesapp0_.id=recipeingr1_.recipe_id 
INNER JOIN recipes_application$ingredient recipesapp2_ ON recipeingr1_.ingredient_id=recipesapp2_.id 
WHERE (recipesapp0_.title like ?) AND (recipesapp2_.name like ?) ORDER BYrecipesapp0_.title ASC

NOTE: Next time when someone asks for clarification of your use-case please give it instead of being unfriendly to them!