我有以下从db返回的对象:
{
"total_rows": 200,
"bookmark": "g1AAAABteJzLYWBgYMpgTmEQTM4vTc5ISXIwNDLXMwBCwxyQVCJDUv3___-zMpjc7D8wgEEiCx71eSwgJQ1A6j-GtiwA6MscCg",
"rows": [
{
"id": "51a1ff51b3b4719d05e40ac4bb0d0566",
"objects": {
"0": {
"type": "ipv4-addr",
"value": "192.168.1.10",
"resolves_to_refs": "2"
},
"1": {
"type": "network-traffic"
}
}
],
"counts": {
"created_by_ref": {
"0203a7e6-b174-4af9-812d-ab889816e868": 1,
"0250789a-14c3-4751-b4a0-c017af82b8f1": 1,
"03c63db6-2a84-4627-88be-a83208d524e6": 1,
"05cba3da-11ff-4a7a-aae9-0b1614cd5300": 1,
"fc825d33-26ea-4563-9478-2e1887b87112": 1
},
"file.hashes.MD5": {
"UNDEFINED": 200
},
"file.name": {
"UNDEFINED": 200
},
"ipv4_addr.value": {
"127.0.0.1": 200,
"192.168.1.10": 200
},
"last_observed": {
"1583503380000": 5,
"1583589780000": 9,
"1585749840000": 12
}
},
"num_of_rows": 10
}我正在尝试将graphql模式应用到上述内容中。我有以下几个可以正常工作的方法:
const graphql = require("graphql");
const { GraphQLObjectType, GraphQLString, GraphQLSchema, GraphQLInt, GraphQLList } = graphql;
const SearchResultType = new GraphQLObjectType({
name: "SearchResult",
fields: ()=>({
total_rows: { type: GraphQLInt },
bookmark: { type: GraphQLString },
//rows: { type: new GraphQLList(GraphQLInt) },
num_of_rows: { type: GraphQLInt }
})
});
const RootQuery = new GraphQLObjectType({
name: "RootQueryType",
fields: {
searchResult:{
type: SearchResultType,
args: { id: { type: GraphQLString } },
resolve(parentValue: any, args: any) {
console.log(args)
return resultMock;
}
}
}
});
module.exports = new GraphQLSchema({
query: RootQuery,
});上述方法适用于已定义的数据类型。但是,在上面的mockResult中有一些对象,例如:
"objects": {
"0": {
"type": "ipv4-addr",
"value": "192.168.1.10",
"resolves_to_refs": "2"
},
"1": {
"type": "network-traffic"
}或
"counts": {
"created_by_ref": {
"0203a7e6-b174-4af9-812d-ab889816e868": 1,
"0250789a-14c3-4751-b4a0-c017af82b8f1": 1,
"03c63db6-2a84-4627-88be-a83208d524e6": 1,
"05cba3da-11ff-4a7a-aae9-0b1614cd5300": 1,
"fc825d33-26ea-4563-9478-2e1887b87112": 1因此,正如你所看到的,这些对象的键是随机的,或者至少在我们收到它们之前是不可猜测的。我有没有办法这样定义一个东西:GraphQLList:{ type: new type (任何我们不知道的随机对象) },作为下面模式中的一个类型:
const SearchResultType = new GraphQLObjectType({
name: "SearchResult",
fields: ()=>({
total_rows: { type: GraphQLInt },
bookmark: { type: GraphQLString },
rows: { type: new GraphQLList(any random object we do not know ) },
num_of_rows: { type: GraphQLInt }
})
});发布于 2020-03-23 06:40:21
您可以使用GraphQL JSON标量(例如,从这个implementation)。但我并不建议这样做(事实上,几年前我做过一次题为"GraphQL JSON标量被认为有害“的演讲)。相反,您可能希望将类似映射的对象转换为键值对列表。
因此,例如,对于counts对象,您可以执行以下操作:
type CreatedByRef {
key: ID
count: Int
}Object.keys(counts.created_by_ref).map(key => ({
key,
count: counts.created_by_ref[key],
}));这将更改结果的形状,但保留GraphQL的所有属性。
https://stackoverflow.com/questions/60795261
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