Question

我想知道是否可以有条件地创建Construct子句的某些部分。例如，假设我们有以下构造查询：

Construct {-:a a <smth:classtype>.
           -:a <smth:attr> ?b} WHERE
          {?c a <smth:anotherCalss>.
           Optional{?c <smth:anotherAttr> ?b}}

在这种情况下，？b并不总是受限于smth。我只想创建空白节点-：a（如果？b是有界的）。有没有办法在sparql的Construct子句中添加这样的条件？

Answer 1

您可以在WHERE子句中有条件地创建bnode，方法是将其放在OPTIONAL中：

import {
  GraphQLString,
  GraphQLBoolean,
  GraphQLInt,
  GraphQLList,
  GraphQLNonNull,
  GraphQLObjectType,
} from 'graphql';

// App Imports
import PostType from '../type';
import { getAll, getById } from '../resolvers';

const ListPostType = new GraphQLObjectType({
  name: 'listpost',
  description: '...',
  fields: () => ({
    rows: {
      type: new GraphQLList(PostType),
    },
    count: {
      type: GraphQLInt,
    },
  }),
});

// Post All
export const posts = {
  type: ListPostType,
  args: {
    first: {
      type: GraphQLInt,
    },
    offset: {
      type: GraphQLInt,
    },
    titulo: {
      type: GraphQLString,
    },
    mes: {
      type: GraphQLInt,
    },
    dia: {
      type: GraphQLInt,
    },
    contenido: {
      type: GraphQLString,
    },
    sort: {
      type: GraphQLString,
    },
    destacado: {
      type: GraphQLBoolean,
    },
    order: {
      type: GraphQLString,
    },
    categoria: {
      type: GraphQLInt,
    },
    parent: {
      type: GraphQLInt,
    },
    tag: {
      type: GraphQLInt,
    },
    equipo: {
      type: GraphQLInt,
    },
  },
  resolve: getAll,
};

// Post By ID
export const post = {
  type: PostType,
  args: {
    id: { type: new GraphQLNonNull(GraphQLInt) },
  },
  resolve: getById,
};

SPARQL中的条件构造

1 个答案: