【发布时间】:2019-10-28 17:16:05
【问题描述】:
如何使用 GQL 指令实现“else”效果
有一种方法可以通过gql 使用@include(if: $withFriends) 有条件地获取某些内容
query Hero($episode: Episode, $withFriends: Boolean!) {
hero(episode: $episode) {
name
friends @include(if: $withFriends) {
name
}
}
}
但如果$withFriends 为假,我想获取其他内容。我可以通过传递额外的变量$notWithFriends来实现它
query Hero($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?
类似这样的:@include(else: $withFriends) 或 @include(ifNot: $withFriends) 或 @include(if: !$withFriends)
【问题讨论】: