graphql-compose-mongoose - graphql-compose-mongoose 中的全文 mongodb $text 搜索查询

Question

我无法弄清楚如何构建 graphql 查询来使用文本索引执行 mongodb 全文搜索。 https://docs.mongodb.com/manual/text-search/

我已经在 mongoose 架构中的字符串上创建了一个文本索引，但在 grapqhl Playground 中显示的架构中没有看到任何内容。

Answer 1

有点晚了，尽管我能够像这样实现

const FacilitySchema: Schema = new Schema(
    {
        name: { type: String, required: true, maxlength: 50, text: true },
        short_description: { type: String, required: true, maxlength: 150, text: true },
        description: { type: String, maxlength: 1000 },
        location: { type: LocationSchema, required: true },
    },
    {
        timestamps: true,
    }
);

FacilitySchema.index(
    {
        name: 'text',
        short_description: 'text',
        'category.name': 'text',
        'location.address': 'text',
        'location.city': 'text',
        'location.state': 'text',
        'location.country': 'text',
    },
    {
        name: 'FacilitiesTextIndex',
        default_language: 'english',
        weights: {
            name: 10,
            short_description: 5,
            // rest fields get weight equals to 1
        },
    }
);

为模型创建 ObjectTypeComposer 后，添加此

const paginationResolver = FacilityTC.getResolver('pagination').addFilterArg({
    name: 'search',
    type: 'String',
    query: (query, value, resolveParams) => {
        resolveParams.args.sort = {
            score: { $meta: 'textScore' },
        };
        query.$text = { $search: value, $language: 'en' };
        resolveParams.projection.score = { $meta: 'textScore' };
    },
});

FacilityTC.setResolver('pagination', paginationResolver);

然后你可以像这样分配

const schemaComposer = new SchemaComposer();

schemaComposer.Query.addFields({
   // ...
   facilities: Facility.getResolver('pagination')
   // ...
});

在您的客户端，像这样执行查询

{
  facilities(filter: { search: "akure" }) {
    count
    items {
      name
    }
  } 
}