java – 为什么在此GraphQL设置中没有调用DataFetcher？

我想编写一段代码,它将处理像这样的GraphQL查询：

query {
      group(id: "com.graphql-java")
      name(name: "graphql-java")
      version(id: "2.3.0")
  }

我创建了一个数据提取器,并在get方法中放置一个断点：

import graphql.schema.DataFetcher;
  import graphql.schema.DataFetchingEnvironment;

  public class TestDataFetcher implements DataFetcher {
      public Object get(final DataFetchingEnvironment dataFetchingEnvironment) {
          return null;
      }
  }

然后我写了以下代码：

public class Example02 {
      public static void main(final String[] args) throws IOException {
          final Example02 app = new Example02();
          app.run();
      }
      void run() throws IOException {
          final TestDataFetcher testDataFetcher = new TestDataFetcher();

          final List<GraphQLFieldDefinition> fields = Lists.newArrayList(
                  createGroupField(testDataFetcher),createNameField(),createVersionField());

          final GraphQLObjectType queryType = newObject()
                  .name("query")
                  .fields(fields)
                  .build();
          final GraphQLSchema schema = GraphQLSchema.newSchema()
                  .query(queryType)
                  .build();
          final String query = FileUtils.readFileToString(
                  new File("src/main/resources/query1.txt"),"UTF-8"
          );
          final Map<String,Object> result = (Map<String,Object>) new GraphQL(schema).execute(query).getData();
          System.out.println(result);
      }

      private GraphQLFieldDefinition createVersionField() {
          return newFieldDefinition().type(GraphQLString).name("version").build();
      }

      private GraphQLFieldDefinition createNameField() {
          return newFieldDefinition().type(GraphQLString).name("name").build();
      }

      private GraphQLFieldDefinition createGroupField(TestDataFetcher testDataFetcher) {
          final GraphQLArgument idArg = newArgument().name("id").type(GraphQLString).build();
          return newFieldDefinition()
                  .type(GraphQLString)
                  .name("group")
                  .dataFetcher(testDataFetcher)
                  .argument(idArg)
                  .build();
      }
  }

当我在调试模式下运行main方法时,断点不被激活.

为什么？我该怎么解决？

解决方法

这是您的工作示例.
我已经使用你发布的查询文件.
根据需要进一步实现您的dataFetcher.
基本上你应该定义名称和版本字段的参数.
当您运行新的GraphQL(架构).execute(query)时,调试器会告诉所有内容
它具有包含所有问题的错误数组.

import graphql.GraphQL;
import graphql.schema.*;
import org.apache.commons.io.FileUtils;

import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

import static graphql.Scalars.GraphQLString;
import static graphql.schema.GraphQLArgument.newArgument;
import static graphql.schema.GraphQLFieldDefinition.newFieldDefinition;
import static graphql.schema.GraphQLObjectType.newObject;

public class Example2 {


    public class TestDataFetcher implements DataFetcher {
        public Object get(DataFetchingEnvironment environment) {
            String id = (String)environment.getArgument("id");
            return id;
        }
    }

    public static void main(final String[] args)  {
        Example2 app = new Example2();
        app.run();
    }
    void run() {
         TestDataFetcher testDataFetcher = new TestDataFetcher();   

         List<GraphQLFieldDefinition> fields = new ArrayList<GraphQLFieldDefinition>();

            fields.add(createGroupField(testDataFetcher));
                fields.add(createNameField());
                fields.add(createVersionField());

         GraphQLObjectType queryType = newObject()
                .name("query")
                .fields(fields)
                .build();

         GraphQLSchema schema = GraphQLSchema.newSchema()
                .query(queryType)
                .build();
        String query = null;
        try {
              query = FileUtils.readFileToString(
                    new File("src/main/resources/query1.txt"),"UTF-8"
            );
        }catch(IOException ioe){
            ioe.printStackTrace();
        }

        if(query!=null) {
            Map<String,Object>) new GraphQL(schema).execute(query).getData();
            System.out.println(result);
        }
    }

    private GraphQLFieldDefinition createVersionField() {
         GraphQLArgument arg = newArgument().name("id").type(GraphQLString).build();
        return newFieldDefinition().type(GraphQLString).name("version").argument(arg).build();
    }

    private GraphQLFieldDefinition createNameField() {
        GraphQLArgument arg = newArgument().name("name").type(GraphQLString).build();
        return newFieldDefinition().type(GraphQLString).name("name").argument(arg).build();
    }

    private GraphQLFieldDefinition createGroupField(TestDataFetcher testDataFetcher) {
        final GraphQLArgument idArg = newArgument().name("id").type(GraphQLString).build();
        return newFieldDefinition()
                .type(GraphQLString)
                .name("group")
                .dataFetcher(testDataFetcher)
                .argument(idArg)
                .build();
    }
}

原文链接：https://www.f2er.com/java/124305.html

java – 为什么在此GraphQL设置中没有调用DataFetcher？

解决方法

猜你在找的Java相关文章