使用 GraphQL 的“其他”

Question

如何使用 GQL 指令实现“else”效果

有一种方法来获取与有条件的东西GQL使用@include(if: $withFriends)

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

但如果$withFriends是假的，我想获取其他东西。我可以通过传递额外的变量来实现它$notWithFriends

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

问题：是否可以避免使用附加变量？

像这样：@include(else: $withFriends)或@include(ifNot: $withFriends)或@include(if: !$withFriends)

Answer 1

您可以使用@skip指令，它的工作原理与@include- 如果if参数为true：则省略选择集中的字段：

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

这样，如果$withFriends是true，只有name和friends会被选中。如果是false，则只有name和appearsIn将被选中。