@include（否则为@include（如果不是，则为：graphql

Question

如何使用GQL指令实现“其他”效果

有一种方法可以使用@include(if: $withFriends)使用gql有条件地获取内容

query Hero($episode: Episode, $withFriends: Boolean!) {
  hero(episode: $episode) {
    name
    friends @include(if: $withFriends) {
      name
    }
  }
}

但是，如果$withFriends为假，我想获取其他信息。我可以通过传递附加变量$notWithFriends

来实现

query Hero($episode: Episode, $withFriends: Boolean!) {
  hero(episode: $episode) {
    name
    friends @include(if: $withFriends) {
      name
    }
    appearsIn @include(if: $notWithFriends)
  }
}

问题：是否可以避免使用其他变量？

类似这样的内容：@include(else: $withFriends)或@include(ifNot: $withFriends)或@include(if: !$withFriends)

Answer 1

您可以使用@skip伪指令，该伪指令的工作方式与@include相反—如果if参数为true，则它会忽略选择集中的字段：< / p>

query Hero($episode: Episode, $withFriends: Boolean!) {
  hero(episode: $episode) {
    name
    friends @include(if: $withFriends) {
      name
    }
    appearsIn @skip(if: $withFriends)
  }
}

通过这种方式，如果$withFriends为true，则只会选择name和friends。如果为false，则只会选择name和appearsIn。