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Code Issues Packages Projects Releases Wiki Activity

Use searchIssueQuery() in countIssueGroupByLabels().

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takezoe
2013-06-27 16:04:46 +09:00
parent 54ce1d720d
commit 2dfb444bf8
1 changed files with 1 additions and 10 deletions
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11
src/main/scala/service/IssuesService.scala
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@@ -60,22 +60,13 @@ trait IssuesService {
def countIssueGroupByLabels(owner: String, repository: String, condition: IssueSearchCondition,
filter: String, userName: Option[String]): Map[String, Int] = {
Issues
searchIssueQuery(owner, repository, condition.copy(labels = Set.empty), filter, userName)
.innerJoin(IssueLabels).on { (t1, t2) =>
(t1.userName is t2.userName) && (t1.repositoryName is t2.repositoryName) && (t1.issueId is t2.issueId)
}
.innerJoin(Labels).on { case ((t1, t2), t3) =>
(t2.userName is t3.userName) && (t2.repositoryName is t3.repositoryName) && (t2.labelId is t3.labelId)
}
.filter { case ((t1, t2), t3) =>
(t1.userName is owner.bind) &&
(t1.repositoryName is repository.bind) &&
(t1.closed is (condition.state == "closed").bind) &&
(t1.milestoneId is condition.milestoneId.get.get.bind, condition.milestoneId.flatten.isDefined) &&
(t1.milestoneId isNull, condition.milestoneId == Some(None)) &&
(t1.assignedUserName is userName.get.bind, filter == "assigned") &&
(t1.openedUserName is userName.get.bind, filter == "created_by")
}
.groupBy { case ((t1, t2), t3) =>
t3.labelName
}