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在我的数据模型中，挑战和它的白名单项目之间存在非常简单的一对多关系。

我正在尝试选择一个按白名单过滤的挑战。基本上，挑战选择标准是当挑战本身在白名单中没有任何条目或白名单按名称匹配时。

这可以通过非常简单的 SQL 查询来实现：

select c.* from challenge c, challenge_whitelist w where (c.id = w."challengeId" and w."userName" = 'testuser') or ((select count(*) where c.id = w."challengeId") = 0);

我无法将其翻译为Exposed：

  // will not compile
  fun listAll(userName: String) {
    ExposedChallenge.wrapRows(
      ChallengeTable.innerJoin(ChallengeWhitelistTable)
        .slice(ChallengeTable.columns)
        .select((ChallengeWhitelistTable.userName eq userName) or (ChallengeTable.innerJoin(ChallengeWhitelistTable).selectAll().count() eq 0))
    ).toList()
  }

用户名检查工作正常，但没有ChallengeTable.innerJoin(ChallengeWhitelistTable).selectAll().count() eq 0)被限定为有效表达式（不会编译）。

请注意，映射非常简单：

object ChallengeTable : IntIdTable() {
  val createdAt = datetime("createdAt")
}

class ExposedChallenge(id: EntityID<Int>) : IntEntity(id) {
  companion object : IntEntityClass<ExposedChallenge>(ChallengeTable)
  var createdAt by ChallengeTable.createdAt
  val whitelist by ExposedChallengeWhitelist referrersOn ChallengeWhitelistTable.challenge
}

object ChallengeWhitelistTable : IntIdTable(name = "challenge_whitelist") {
  var userName = varchar("userName", 50)
  var challengeId = integer("challengeId")
  val challenge = reference("challengeId", ChallengeTable).uniqueIndex()
}

class ExposedChallengeWhitelist(id: EntityID<Int>) : IntEntity(id) {
  companion object : IntEntityClass<ExposedChallengeWhitelist>(ChallengeWhitelistTable)

  val challengeId by ChallengeWhitelistTable.challengeId
  val challenge by ExposedChallenge referencedOn ChallengeWhitelistTable.challenge
}

任何帮助，将不胜感激。

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1 回答 1

2

Your SQL query is invalid as you use select count(*) without from part. But it can be rewritten with Exposed DSL like:

ChallengeTable.leftJoin(ChallengeWhitelistTable).
    slice(ChallengeTable.columns).
    selectAll().
    groupBy(ChallengeTable.id, ChallengeWhitelistTable.userName).having {
    (ChallengeWhitelistTable.userName eq "testUser") or
    (ChallengeWhitelistTable.id.count() eq 0)
}

Another way is to use just left join:

ChallengeTable.leftJoin(ChallengeWhitelistTable).
    slice(ChallengeTable.columns).
    select {
        (ChallengeWhitelistTable.userName eq "testUser") or
        (ChallengeWhitelistTable.id.isNull())
    }
于 2019-09-24T19:11:16.207 回答