1. 概述
在本系列文章的第一篇中,我们将探索如何为 REST API 构建一个简单的查询语言。后端我们使用 Spring,数据访问层则采用 JPA 2 的 Criteria API 来实现查询逻辑。
为什么要引入查询语言?
因为对于稍微复杂一点的 API 来说,仅仅通过简单的字段进行资源过滤是远远不够的。一个设计良好的查询语言可以提供更高的灵活性,让你更精准地筛选出需要的资源。
2. User 实体类
我们以一个简单的 User
类作为示例实体类,用于后续的查询操作:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String firstName;
private String lastName;
private String email;
private int age;
}
3. 使用 JPA Criteria 构建动态查询
现在进入核心部分:如何在持久层构建灵活的查询逻辑。
构建查询抽象层的关键在于平衡灵活性和复杂度。我们希望提供足够的查询能力,同时不让代码变得难以维护。
核心逻辑是:传入一组查询条件,返回匹配的结果集。
来看一个示例实现:
@Repository
public class UserDAO implements IUserDAO {
@PersistenceContext
private EntityManager entityManager;
@Override
public List<User> searchUser(List<SearchCriteria> params) {
CriteriaBuilder builder = entityManager.getCriteriaBuilder();
CriteriaQuery<User> query = builder.createQuery(User.class);
Root<User> root = query.from(User.class);
Predicate predicate = builder.conjunction();
UserSearchQueryCriteriaConsumer searchConsumer =
new UserSearchQueryCriteriaConsumer(predicate, builder, root);
params.forEach(searchConsumer);
predicate = searchConsumer.getPredicate();
query.where(predicate);
return entityManager.createQuery(query).getResultList();
}
@Override
public void save(User entity) {
entityManager.persist(entity);
}
}
接着看 UserSearchQueryCriteriaConsumer
类的实现:
public class UserSearchQueryCriteriaConsumer implements Consumer<SearchCriteria> {
private Predicate predicate;
private CriteriaBuilder builder;
private Root<User> root;
public UserSearchQueryCriteriaConsumer(Predicate predicate, CriteriaBuilder builder, Root<User> root) {
this.predicate = predicate;
this.builder = builder;
this.root = root;
}
@Override
public void accept(SearchCriteria param) {
if (param.getOperation().equalsIgnoreCase(">")) {
predicate = builder.and(predicate, builder
.greaterThanOrEqualTo(root.get(param.getKey()), param.getValue().toString()));
} else if (param.getOperation().equalsIgnoreCase("<")) {
predicate = builder.and(predicate, builder.lessThanOrEqualTo(
root.get(param.getKey()), param.getValue().toString()));
} else if (param.getOperation().equalsIgnoreCase(":")) {
if (root.get(param.getKey()).getJavaType() == String.class) {
predicate = builder.and(predicate, builder.like(
root.get(param.getKey()), "%" + param.getValue() + "%"));
} else {
predicate = builder.and(predicate, builder.equal(
root.get(param.getKey()), param.getValue()));
}
}
}
public Predicate getPredicate() {
return predicate;
}
}
最后是 SearchCriteria
查询条件类:
public class SearchCriteria {
private String key;
private String operation;
private Object value;
}
该类包含三个字段:
- ✅
key
:对应实体字段名,如firstName
,age
等 - ✅
operation
:支持的操作符,如:
,<
,>
等 - ✅
value
:对应的值,如john
,25
等
4. 单元测试查询逻辑
我们通过几个单元测试来验证查询逻辑是否正确。
先准备测试数据:
@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(classes = { PersistenceConfig.class })
@Transactional
@TransactionConfiguration
public class JPACriteriaQueryTest {
@Autowired
private IUserDAO userApi;
private User userJohn;
private User userTom;
@Before
public void init() {
userJohn = new User();
userJohn.setFirstName("John");
userJohn.setLastName("Doe");
userJohn.setEmail("john@example.com");
userJohn.setAge(22);
userApi.save(userJohn);
userTom = new User();
userTom.setFirstName("Tom");
userTom.setLastName("Doe");
userTom.setEmail("tom@example.com");
userTom.setAge(26);
userApi.save(userTom);
}
}
测试用例1:按 firstName
和 lastName
查询用户:
@Test
public void givenFirstAndLastName_whenGettingListOfUsers_thenCorrect() {
List<SearchCriteria> params = new ArrayList<>();
params.add(new SearchCriteria("firstName", ":", "John"));
params.add(new SearchCriteria("lastName", ":", "Doe"));
List<User> results = userApi.searchUser(params);
assertThat(userJohn, isIn(results));
assertThat(userTom, not(isIn(results)));
}
测试用例2:只按 lastName
查询:
@Test
public void givenLast_whenGettingListOfUsers_thenCorrect() {
List<SearchCriteria> params = new ArrayList<>();
params.add(new SearchCriteria("lastName", ":", "Doe"));
List<User> results = userApi.searchUser(params);
assertThat(userJohn, isIn(results));
assertThat(userTom, isIn(results));
}
测试用例3:按 age > 25
查询:
@Test
public void givenLastAndAge_whenGettingListOfUsers_thenCorrect() {
List<SearchCriteria> params = new ArrayList<>();
params.add(new SearchCriteria("lastName", ":", "Doe"));
params.add(new SearchCriteria("age", ">", "25"));
List<User> results = userApi.searchUser(params);
assertThat(userTom, isIn(results));
assertThat(userJohn, not(isIn(results)));
}
测试用例4:按不存在的字段值查询:
@Test
public void givenWrongFirstAndLast_whenGettingListOfUsers_thenCorrect() {
List<SearchCriteria> params = new ArrayList<>();
params.add(new SearchCriteria("firstName", ":", "Adam"));
params.add(new SearchCriteria("lastName", ":", "Fox"));
List<User> results = userApi.searchUser(params);
assertThat(userJohn, not(isIn(results)));
assertThat(userTom, not(isIn(results)));
}
测试用例5:模糊匹配 firstName
:
@Test
public void givenPartialFirst_whenGettingListOfUsers_thenCorrect() {
List<SearchCriteria> params = new ArrayList<>();
params.add(new SearchCriteria("firstName", ":", "jo"));
List<User> results = userApi.searchUser(params);
assertThat(userJohn, isIn(results));
assertThat(userTom, not(isIn(results)));
}
5. 控制器层对接 REST 接口
最后,我们将查询逻辑接入 REST API,实现一个支持查询语言的接口。
@Controller
public class UserController {
@Autowired
private IUserDao api;
@RequestMapping(method = RequestMethod.GET, value = "/users")
@ResponseBody
public List<User> findAll(@RequestParam(value = "search", required = false) String search) {
List<SearchCriteria> params = new ArrayList<>();
if (search != null) {
Pattern pattern = Pattern.compile("(\\w+?)(:|<|>)(\\w+?),");
Matcher matcher = pattern.matcher(search + ",");
while (matcher.find()) {
params.add(new SearchCriteria(matcher.group(1),
matcher.group(2), matcher.group(3)));
}
}
return api.searchUser(params);
}
}
你可以这样调用接口进行查询:
http://localhost:8082/spring-rest-query-language/auth/users?search=lastName:doe,age%3E25
返回结果如下:
[{
"id":2,
"firstName":"tom",
"lastName":"doe",
"email":"tom@example.com",
"age":26
}]
6. 小结
这个实现虽然还比较粗糙,但已经具备了基本的查询能力。它为我们提供了一个灵活、可扩展的 REST 查询基础。在后续文章中我们会继续优化这个查询语言,比如加入分页、排序、更复杂的逻辑操作等。
完整代码可以在 GitHub 上找到:GitHub 项目地址 ✅
如果你正在构建自己的 REST API,不妨试试这个思路。它简单、灵活、不依赖第三方库,适合中大型项目快速搭建查询功能。