使用GraphQLObjectType的Hello world
使用GraphQLObjectType的Hello world
提问于 2016-04-21 12:11:12
我正在执行下面的代码，结果是：
{TestPojo={id=null, name=null}}我期望得到的结果是{TestPojo={id="1", name="Jack"}}。我遗漏了什么？
import static graphql.Scalars.GraphQLString;
import static graphql.schema.GraphQLFieldDefinition.newFieldDefinition;
import static graphql.schema.GraphQLObjectType.newObject;
import java.util.Map;
import graphql.GraphQL;
import graphql.schema.DataFetcher;
import graphql.schema.DataFetchingEnvironment;
import graphql.schema.GraphQLObjectType;
import graphql.schema.GraphQLSchema;
public class HelloWorld {
public static void main(String[] args) {
// sub schema to be added to parent schema
GraphQLObjectType testPojo = newObject().name("TestPojo")
.description("This is a test POJO")
.field(newFieldDefinition().name("id").type(GraphQLString).build())
.field(newFieldDefinition().name("name").type(GraphQLString).build())
.build();
// parent schema
GraphQLObjectType queryType = newObject().name("helloWorldQuery")
.field(newFieldDefinition().name(testPojo.getName())
.type(testPojo)
.dataFetcher(new DataFetcher() {
@Override
public Object get(DataFetchingEnvironment arg0) {
Object a = new GrapgQLSampleController()
.greeting2();
return a;
}
})
.build())
.build();
GraphQLSchema schema = GraphQLSchema.newSchema().query(queryType).build();
Map<String, Object> result = (Map<String, Object>) new GraphQL(schema).execute("{TestPojo {id,name}}")
.getData();
System.out.println(result);
// Prints: {TestPojo={id=null, name=null}}
}
/**
* service method 2
*
* @return
*/
public TestPojo greeting2() {
return new TestPojo("1", "Jack");
}
// inner pojo
class TestPojo {
public String id;
public String name;
TestPojo(String id, String name) {
this.id = id;
this.name = name;
}
}
}回答 2
Stack Overflow用户
发布于 2017-07-10 18:11:04
尝试在TestPojo类上添加getter方法，因为这些方法在类上声明时将被调用。否则，在现场定义您自己的DataFetcher。
class TestPojo {
public final String id;
public final String name;
TestPojo(String id, String name) {
this.id = id;
this.name = name;
}
String getId() {
return id;
}
String getName() {
return name;
}
}Stack Overflow用户
发布于 2017-01-04 02:15:39
试图通过修复代码来变得更具教育性。请注意，静态内容来自.staticValue中的任何内容(“X”)
GraphQLObjectType testPojo = newObject().name("TestPojo")
.description("This is a test POJO")
.field(newFieldDefinition().type(GraphQLString).name("id").staticValue("Test1"))
.field(newFieldDefinition().type(GraphQLString).name("name").staticValue("pojo1"))
.build();
GraphQLSchema schema = GraphQLSchema.newSchema().query(testPojo).build();
Map<String, Object> result = (Map<String, Object>) new GraphQL(schema).execute("{id,name}")
.getData();
System.out.println(result);将打印{id=Test1，name=pojo1}
当我有时间时，我可以使用DataFetcher更新示例
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/36759503
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