如何编写http请求url以获得类似以下内容的查询：

select *
from incidents i,
     jira_issues ji
where i.incident_id = ji.incident_id
  and ji.external_jira_issue_id = 'ABC-123'
  and ji.jira_server_id = '1'

我有以下课程：

@Entity(name = "incidents")
public class IncidentEntity {
  @OneToMany(
      mappedBy = "incident",
      cascade = CascadeType.ALL
  )
  @LazyCollection(LazyCollectionOption.FALSE)
  private List<JiraIssueEntity> jiraIssues;
...
}

@Entity(name = "jira_issues")
public class JiraIssueEntity {

  @EmbeddedId
  @EqualsAndHashCode.Include
  private JiraIssueId id;

  @ManyToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "incident_id")
  @ToString.Exclude
  private IncidentEntity incident;
...
}

@Embeddable
public class JiraIssueId implements Serializable {

  @EqualsAndHashCode.Include
  private String externalJiraIssueId;
  @EqualsAndHashCode.Include
  private String jiraServerId;
}

这是我的api方法签名：

@GetMapping("")
  public Page<Incident> listIncidents(
      @QuerydslPredicate(root = IncidentEntity.class) Predicate predicate
  );

我知道我可以寄一些东西，比如：

/incidents/?jiraIssues.id.externalJiraIssueId=ABC-123&jiraIssues.id.jiraServerId=1"

这将转换为以下查询：

select *
from incidents incidenten0_
where (exists(select 1
              from jira_issues jiraissues1_
              where incidenten0_.incident_id = jiraissues1_.incident_id
                and (lower(jiraissues1_.external_jira_issue_id) like ? escape '!')))
  and (exists(select 1
              from jira_issues jiraissues2_
              where incidenten0_.incident_id = jiraissues2_.incident_id
                and (lower(jiraissues2_.jira_server_id) like ? escape '!')))

这不太好。
我不知道如何：
do equals和not contains（externaljiraissueid=-1234的行也将返回，但我不希望返回）。
检查相同的jiraissue是否有externaljiraissueid=-123和jiraissues.id.jiraserverid=1，并且没有不同的jiraissues，每个jiraissues都匹配一个（类似jiraissues.id=（-123，1）
谢谢您。