SilverStripe GraphQL - 查询具有后代的类型时出错

Question

我在尝试查询类型和后代时遇到此错误：

"Fragment cannot be spread here as objects of type \"AppTestObject\" can never be of type \"AppTestChild\"."

我使用 recipe-core、admin、graphql 和 graphql-devtools（所有最新版本）设置了测试安装，以在基本设置中对此进行测试。我创建了 2 个对象：

class TestObject extends DataObject {

    private static $singular_name = "Test Object";
    private static $plural_name = "Test Objects";
    private static $table_name = "TestObject";

    private static $db = [
        'Title' => 'Varchar(255)'
    ];

}

class TestChild extends DataObject {

    private static $singular_name = "Test Child";
    private static $plural_name = "Test Children";
    private static $table_name = "TestChild";

    private static $db = [
        'Title' => 'Varchar(255)'
    ];

}

并通过配置搭建简单的脚手架：

SilverStripe\GraphQL\Controller:
  schema:
    scaffolding:
      types:

        App\TestObject:
          fields: [ID]
          operations:
            read: true

        App\TestChild:
          fields: [ID, Title]
          operations:
            read: true

我能够毫无问题地单独查询这些类型中的每一种。但是当我试图让 TestChild 作为 TestObject 的后代时，我得到了上面的错误。这是我的查询示例：

query {
  readAppTestObjects {
    edges {
      node {
        ...on AppTestChild {
          Title
        }
      }
    }
  }
}

检查 graphiql 模式的文档，readAppTestObjects 下没有任何引用后代的内容，尽管在 silverstripe/graphql 的文档中它说：

When reading types that have exposed descendants (e.g. reading Page, when RedirectorPage is also exposed), the return type is a union of the base type and all exposed descendants. This union type takes on the name {BaseType}WithDescendants.

Answer 1

是的，这是 SilverStripe graphql 模块中的错误。你正在做的应该有效。

我相信修复程序正在 https://github.com/silverstripe/silverstripe-graphql/pull/176 进行中，您可以在那里关注进度。也许试试这个补丁并留下一些评论。

SilverStripe GraphQL - 查询具有后代的类型时出错

SilverStripe GraphQL - error when querying types with descendants

php

silverstripe

graphql

silverstripe-4