GraphQL:如何使用 graphQL

问题描述

目前，我在 graphql-java图书馆.它确实有一些基本的中继支持，我们可以在其中创建一个 connection，这是 Facebook 推荐的实现分页的方式.
这个 是帮助实现这一目标的方法.但是，由于没有文档，我发现很难理解这个函数是如何工作的.如果他们已经有一个允许基本查询的现有模型，例如 Add、delete、fetch等使用graphql-java库?

Relay relay = new Relay();
GraphQLOutputType book = ...; //build your normal Book object type
GraphQLObjectType bookEdge = relay.edgeType(book.getName(), book, null, Collections.emptyList());
GraphQLObjectType bookConnection = relay.connectionType(book.getName(), bookEdge, Collections.emptyList());

因此，您将拥有一个符合 BookConnection 类型"nofollow noreferrer">中继连接规范.

As a result, you'd have a BookConnection type that conforms to the Relay connection spec.

至于使用基本 GraphQL 的示例，您有一个简单的网络应用程序here.

As for the example with basic GraphQL, you have a simple web app here.

连接规范自然适合支持基于光标的分页的数据存储，但在使用不同的分页样式时需要一些创造力.

The connection spec naturally fits a data store that supports cursor based pagination, but needs some creativity when used with different pagination styles.

1) 如果你想使用简单的基于偏移量的分页，你可以决定将 after 视为偏移量(意味着将传递一个数字)，并将 first 视为限制:

1) If you wish to use simple offset based paging, you can decide to treat after as the offset (meaning a number would be passed), and first as the limit:

SELECT * FROM ORDER BY timestamp OFFSET $after LIMIT $first

before 和 last 一样，只是方向不同.

The same for before and last, just different direction.

2) 另一种方法是将 after/before 视为排序列最后看到的值(因此将传递实际(混淆)值):

2) Another way is to treat after/before as the last seen value of the sort column (so an actual (obfuscated) value would be passed):

SELECT * FROM ORDER BY timestamp WHERE timestamp > $after LIMIT $first

我还建议您查看我的项目，graphql-spqr，使用示例应用，这使得开发 GraphQL API 变得非常简单.

I'd also recommend you take a look at my project, graphql-spqr, with an example app, that makes developing GraphQL APIs dead simple.

例如，您将创建这样的分页结果:

For example, you'd create a paginated result like this:

public class BookService {
    @GraphQLQuery(name = "books")
    //make sure the argument names and types match the Relay spec
    public Page<Book> getBooks(@GraphQLArgument(name = "first") int first, @GraphQLArgument(name = "after") String after) {
        //if you decide to fetch from a SQL DB, you need the limit and offset instead of a cursor
        //so, you can treat "first" as count as "after" as offset
        int offset = Integer.valueOf(after);
        List<Book> books = getBooksFromDB(first, offset);
        Page<Book> bookPage = PageFactory.createOffsetBasedPage(books, totalBookCount, offset);
        return bookPage;
    }
}

还有很多其他方法可以创建 Page 实例，这只是最直接的一种.

There's many other ways to create a Page instance, this is just the most straight-forward one.

然后您将从您的 Java 类生成一个架构:

You'd then generate a schema from your Java class:

GraphQLSchema schema = new GraphQLSchemaGenerator()
       .withOperationsFromSingleton(new BookService())
       .generate();
GraphQL graphQL = GraphQLRuntime.newGraphQL(schema).build();

并执行查询:

ExecutionResult result = graphQL.execute("{books(first:10, after:"20") {" +
                "   pageInfo {" +
                "       hasNextPage" +
                "   }," +
                "   edges {" +
                "       cursor, node {" +
                "           title" +
                "}}}}");

但是，再说一次，如果您不使用 Relay，那么真的没有必要让事情变得过于复杂.如果您的存储自然支持基于游标的分页，那就去吧.如果没有，只需使用简单的限制/偏移参数并返回一个列表，而忘记连接规范.它的创建是为了让 Relay 能够在各种情况下自动管理分页，因此如果您不使用 Relay 和/或基于游标的分页的数据库，它几乎总是过度杀伤.

But, again, if you are not using Relay there's really no need to overcomplicate things. If your storage supports cursor-based pagination naturally, go for it. If it doesn't, just use the simple limit/offset arguments and return a list, and forget the connection spec. It was created to enable Relay to automatically manage paging in various scenarios, so it's almost always a total overkill if you're not using Relay and/or a DB with cursor-based pagination.

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