Question

我们如何使spring-data-rest为查询返回单个资源，而不是嵌入列表？这将使客户端的导航更加直观。

e.g。

t.follow("search", "byNameAndType", "identity", "mainContact")

而不是

t.follow("search", "byNameAndType", "$_embedded.credentials[0]._links.identity.href", "mainContact")

我有一个总是返回1或0结果的JPA存储库。唯一性由数据库约束强制执行。

@RepositoryRestResource
public interface CredentialRepository extends PagingAndSortingRepository<Credential, Long>,
    ExternalIdRepository<Credential, Long> {


@RestResource(path = "byNameAndType", rel = "byNameAndType")
Credential findByNameIgnoreCaseAndTypeIgnoreCase(@Param("name") String name, @Param("type") String type);

}

spring-data-rest而不是返回单个凭证对象，它返回一个嵌入列表。

$ curl -H "Accept: text/plain, application/hal+json, */*" -H "X-VCIDB-User-Id:testUser" http://127.0.0.1:8090/credentials/search/byNameAndType?name=V6UqkSG8\&type=myType
{
  "_embedded" : {
"credentials" : [ {
  "version" : 1,
  "lastUpdTs" : "2014-11-13T12:08:49.301+13:00",
  "lastUpdBy" : ":integration-test",
  "createdTs" : "2014-11-13T12:08:49.092+13:00",
  etc etc

相反，它应该返回相同，就像我通过主键检索它一样。

{
  "version" : 1,
  "lastUpdTs" : "2014-11-13T12:08:49.301+13:00",
  etc etc

否则客户端的API看起来很糟糕

    Traverson t = new Traverson(new URI("http://127.0.0.1:8090/credentials"), MediaTypes.HAL_JSON);
    t.setRestOperations(template);
    Map<String, Object> params = new HashMap<>();
    params.put("name", "V6UqkSG8");
    params.put("type", "myType");
    String contactUuid = t.follow("search", "byNameAndType", "$_embedded.credentials[0]._links.identity.href", "mainContact")
            .withTemplateParameters(params).<String> toObject("$.uuid");
    LOG.info(contactUuid);
    assertThat(contactUuid, Matchers.is("7a7faeaf-6da3-4188-9d28-afbb30ce38b3"));

当客户端的transversion更容易理解时：

String contactUuid = t.follow("search", "byNameAndType", "identity", "mainContact")
            .withTemplateParameters(params).<String> toObject("$.uuid");

如何使用spring数据REST从查询中返回单个项目

0 个答案: