Question

我已将作者定义如下：

const Author = new GraphQLObjectType({
  name: 'Author',
  description: 'Represent the type of an author of a blog post or a comment',
  fields: () => ({
    _id: {type: GraphQLString},
    name: {type: GraphQLString},
    posts: {type: Post}
  })
});

我已将我的查询定义如下

const Query = new GraphQLObjectType({
  name: "Root_Query",
  fields: {
    authors: {
      type: new GraphQLList(Author),
      args: {_id: { type: GraphQLString },name:{type: GraphQLString}}, 
      resolve: function(rootValue, args, info) {
        let fields = {};
        let fieldASTs = info.fieldASTs;
        fieldASTs[0].selectionSet.selections.map(function(selection) {
          fields[selection.name.value] = 1;
        });
        return authorsCollection.find({}, fields).toArray();
      }
    }

现在，当我使用下面给出的查询查询具有特定ID的作者时

{
  authors(_id: "57c5794a92aef65040c4e0e6"){
    _id
    name
  }
}

而不是使用_id 57c5794a92aef65040c4e0e6显示作者。它显示所有作者_id和name。我该如何解决这个问题？

Answer 1

问题在于以下数据库代码，由于空查询{}而获取所有作者：

return authorsCollection.find({}, fields).toArray()

将_id添加到您的查询中：

// import {ObjectID} from 'mongodb';
const authorId = ObjectID.createFromHexString(args._id);
return authorsCollection.find({_id: authorId}, fields).toArray()

你必须

我正在查询作者特别是＆＃39; _id＆＃39;但它显示所有作者[Graphql]

1 个答案: