阿波罗GraphQL团队说,readQuery和writeQuery对95%的用例都很好。我正在使用useMutation和update,并且希望在不需要调用refetchQueries的情况下从缓存中删除一个项。我的代码如下:
const [deleteSpeaker] = useMutation(DELETE_SPEAKER, {
update(cache, { data: {deleteSpeaker}}) {
const { speakers} = cache.readQuery({query: GET_SPEAKERS});
cache.writeQuery({
query: GET_SPEAKERS,
data: { speakers: speakers.filter(speaker => speaker.id !== deleteSpeaker.id) }
});
},
});从readQuery返回的内容使我认为我应该对speakers.datalist进行过滤,但是当我这样做时,缓存不会更新。
更新缓存以反映从GET_SPEAKERS查询中删除的记录的正确方法是什么。
export const DELETE_SPEAKER = gql`
mutation DeleteSpeaker($speakerId: Int!) {
deleteSpeaker(speakerId: $speakerId) {
id
first
last
favorite
}
}
`;和GET_SPEAKERS
export const GET_SPEAKERS = gql`
query {
speakers {
datalist {
id
first
last
favorite
company
}
}
}
`;发布于 2022-08-09 06:24:48
阅读阿波罗博士,这应该是类似的东西:
const [deleteSpeaker] = useMutation(DELETE_SPEAKER, {
update(cache, { data: {deleteSpeaker}}) {
cache.modify({
id: cache.identify(deleteSpeaker.id),
fields: {
comments(existingSpeakerRefs, { readField }) {
return existingSpeakerRefs.filter(
speaker => deleteSpeaker.id !== readField('id', speakerRef)
);
},
},
});
},
});https://stackoverflow.com/questions/64344886
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